John NachbarWashington University in St. LouisFebruary 15, 2018

Metric Spaces1

1 Metric Spaces Basics.

1.1 Metric spaces.

A metric space (X, d) consists of a set of points, X, together with a function d : X×X → R+, called a metric. The interpretation is that d(x, y) is the distance betweenx and y. To qualify as a metric, the function d must satisfy certain properties.

Definition 1. Given a set X, a function d : X ×X → R+ is a metric iff, for anyx, y, z ∈ X, the following properties hold.

1. d(x, y) = 0 iff x = y.

2. d(x, y) = d(y, x).

3. d(x, z) ≤ d(x, y) + d(y, z).

The third property is called the triangle inequality.

The triangle inequality gets its name because, in standard geometry, if the threepoints x, y, and z form a triangle then the length of the side from x to z is less thenthe sum of the lengths of the other two sides.

Think of X as fundamental while the metric d is an overlay, like the latitude andlongitude grid on a map. Any X has an infinity of possible metrics. At a minimum,metrics can differ because of units (inches rather than centimeters). But it is alsopossible for different metrics to give different answers to questions like, “which isgreater, the distance from x to y or the distance from x to z?” Which metric Ichoose depends entirely on which is most helpful to me. For some spaces, notablyX = RN , there is a default metric that is convenient for almost any application.For other spaces, such as variants of X = Rω, there is no default metric.

Let (X, d) be a metric space and let A ⊆ X. Unless I state explicitly otherwise,I assume that A inherits the d metric, with domain restricted to A × A. A is thena metric subspace of X.

1cbna. This work is licensed under the Creative Commons Attribution-NonCommercial-ShareAlike 4.0 License.

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1.2 Metrics derived from norms.

Many, although not all, of the metric spaces used in economics are derived fromnormed vector spaces. Any norm f induces a metric via df (x, y) = f(x− y).

Theorem 1. Let X be a normed vector space, with norm f . Define df : X×X → R+

by df (x, y) = f(x− y). Then df is a metric.

Proof. I must check that the three metric properties hold.

1. df (x, y) = 0 iff x = y, since f(x− y) = 0 iff x− y = 0 iff x = y.

2. df (x, y) = df (y, x) since f(x− y) = | − 1|f(x− y) = f(y − x).

3. df (x, z) ≤ df (x, y) + df (y, z) since f(x− z) = f(x− y + y − z) ≤ f(x− y) +f(y − z).

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In particular, if df is derived from a norm f then df (x, 0) = f(x). In general,when dealing with a normed vector space, it is implicit that the metric used is theone derived from the norm.

1.3 Examples of Metric Spaces.

1.3.1 RN with the Euclidean metric.

Recall that if x ∈ RN then the Euclidean norm of x is

‖x‖ = (x · x)1/2 .

If N = 1 then ‖x‖ = |x|. The notes on RN established that ‖x‖ does indeed satisfythe required norm properties.

The Euclidean metric is then defined by,

d(x, y) = ‖x− y‖ = [(x− y) · (x− y)]1/2 =

√∑n

(xn − yn)2.

By Theorem 1, d is a metric.Whenever I write RN , you can assume that the metric is the Euclidean metric

unless I explicitly state otherwise. The Euclidean metric is also the default whenI work with subsets of RN , such as the interval (0, 1] or Q (the set of rationalnumbers).

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1.3.2 RN with the max metric.

Although d is the default metric for RN , there are many (infinitely many) otherpossible metrics. Recall the max norm from the notes on normed vector spaces.

‖x‖max = maxn|xn|.

The notes on normed vector spaces verify that the max norm is indeed a norm.Define the max metric as,

dmax(x, y) = ‖x− y‖max = maxn|xn − yn|.

Since, the max norm is a norm, Theorem 1 implies that dmax is, indeed, a metric.

1.3.3 RN with the discrete metric.

For x, y ∈ RN , define

dd(x, y) =

{1 if x 6= y,

0 if x = y.

This is called the discrete metric. It is easy to verify that it is a true metric. It isnot, however, generated by any norm and has a number of peculiar properties. It isuseful mainly for counterexamples.

1.3.4 `∞ with the sup metric.

Recall, from the notes on normed vector spaces, that Rω is the set of points of theform (x1, x2, x3, . . . ), where each xn ∈ R. Also recall that `∞ is the set of points inRω that are bounded in the sense that x ∈ `∞ iff there as an M (which can dependon x) such that for all n, |xn| < M . For x ∈ `∞, the sup norm, given by,

‖x‖sup = supn|xn|

is well defined. The notes on normed vector spaces establish that `∞ is a vectorspace and that ‖ · ‖sup is indeed a norm.

Define the sup metric as,

dsup(x, y) = ‖x− y‖sup = supn|xn − yn|.

By Theorem 1, dsup is, indeed, a metric.

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1.3.5 Rω with the dpw metric.

For any x, y ∈ Rω, define

dpw(x, y) = supn

min{1, |xn − yn|}n

.

(This metric has no standard name. I am calling it dpw where the “pw” is mnemonicfor “pointwise;” the reason for this name is addressed in the notes on Completenessand Compactness in Rω.)

Loosely, there are two differences between dpw and dsup. First, dpw weights high-numbered coordinates less than low numbered coordinates. Second, dpw truncatesthe coordinate by coordinate measure of distance to a maximum value of 1. Becauseof the truncation, dpw, unlike dsup is defined, on all of Rω.

Theorem 2. dpw is a metric on Rω.

Proof. Exercise. �

dpw is not generated by a norm. In particular, the norm would have to bef(x) = dpw(x, 0). But this f violates norm property 2. For example, if x = (1, 1, . . . )and y = (2, 2, . . . ), then f(x) = f(y) = 1.

2 Balls.

Let (X, d) be a metric space. For ε ∈ R, ε > 0, the open ε ball around x is

Nε(x) = {a ∈ X : d(x, a) < ε}.

Think of Nε as mnemonic for “nearby within ε.”

Example 1. For the Euclidean metric on RN . In R, Nε(x) is simply the interval(x− ε, x+ ε). In R2, for the Euclidean metric, Nε(x) is a disk of radius ε, centeredat x, excluding the boundary circle. In R3, Nε(x) is a solid ball of radius ε, centeredat x, excluding the boundary sphere. �

Example 2. For the Euclidean metric on [0, 1] ⊆ R, if ε < 1, Nε(0) = [0, ε), which Iwill also write as (−ε, ε) ∩ R, with (−ε, ε) understood to denote the full interval inR of length 2ε. �

Example 3. For the Euclidean metric on Q, Nε(0) = (−ε, ε)∩Q, again with (−ε, ε)understood to denote the full interval in R of length 2ε. �

Example 4. For the max metric on R2, Nε(x) is the interior of a square with sides oflength 2ε, centered at x. In R3, Nε(x) is the interior of a cube with sides of length2ε, centered at x. �

Example 5. For the discrete metric on R, for any x ∈ R, if ε ≤ 1 then Nε(x) = {x}while if ε > 1 then Nε(x) = R. �

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Remark 1. In a vector space X, a set A ⊆ X is convex iff for any two points x, y ∈ Aand any θ ∈ [0, 1], θx+ (1− θ)y ∈ A; in words, a set is convex iff for any two pointsin A, the line segment joining them is also in A. , Recall (from the notes on normedvector spaces) that a function f : X → R is convex iff for any x, y ∈ X, and anyθ ∈ [0, 1], f(θx + (1 − θ)y) ≤ θf(x) + (1 − θ)f(y). A simple example of a convexfunction is absolute value on R: f(x) = |x|.

It is easy to prove that if a function is convex then for any b ∈ R, the set{x ∈ X : f(x) < b} (which may be empty) is convex. In the case of a metric basedon a norm, this implies that, since any norm is a convex function (see the notes onnormed vector spaces), any ε ball is convex. �

3 Bounded Sets and Totally Bounded Sets.

3.1 Basic Definitions.

Let (X, d) be a metric space. A ⊆ X is bounded iff there is an x ∈ X and r > 0,such that A ⊆ Nr(x). In words, A is bounded iff it is contained in some ball of largeenough radius. In a metric vector space, one can always take x to be the origin.2

A set A is totally bounded iff for any ε > 0 there is a finite set of points E ⊆ Xsuch that

A ⊆⋃x∈E

Nε(x).

If A ⊆⋃x∈E Nε(x) then the set of balls {Nε(x)}x∈E is said to cover A. In words,

then, A is totally bounded iff, for any ε > 0 no matter how small, A can be coveredby a finite number of open balls of radius ε (the number of balls required willtypically go up as ε shrinks).

I leave it as an exercise to show that A is totally bounded iff for any ε > 0there is a finite set E ⊆ A such that {Nε(x)}x∈E covers A (the difference being thatE ⊆ A rather than E ⊆ X).

The next result establishes that any totally bounded set is bounded.

Theorem 3. Let (X, d) be a metric space. If A ⊆ X is totally bounded then it isbounded.

Proof. Since A is totally bounded there is a finite set E ⊆ X such that A is coveredby the balls N1(x) for x ∈ E. Fix any x∗ ∈ X and let r = maxx∈E d(x∗, x) + 1.I claim that for any a ∈ A, a ∈ Nr(x

∗), which shows that A is bounded. Forany a ∈ A, there is an xa ∈ E such that a ∈ N1(xa). By the triangle inequality,d(x∗, a) ≤ d(x∗, xa) + d(xa, a) < d(x∗, xa) + 1 ≤ r. �

2Suppose that A ⊆ Nr(x). Let r̂ = r + d(x, 0). Then, by the triangle inequality, for any a ∈ A,d(a, 0) ≤ d(a, x) + d(x, 0) < r + d(x, 0) = r̂. Therefore, A ⊆ Nr̂(0).

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The converse of Theorem 3 is true in RN , as recorded in Section 3.2 (the proofis deferred to Section 6), but not in general, as illustrated in Section 3.3.

Remark 2. It is not hard to show that a set A is totally bounded iff for any ε > 0,A can be covered by a finite set of ε balls that have centers in A (rather than, as inthe definition, centers in X). �

3.2 Bounded and Totally Bounded Sets in RN .

Theorem 4. If A ⊆ RN is bounded then it is totally bounded.

Proof. Deferred to Section 6. �

As illustrated in Section 3.3, in metric spaces other than RN , bounded sets maynot be totally bounded. The proof of Theorem 4 in Section 6 hints at what can gowrong: for a fixed r, the number of open ε balls used to cover an open r ball in RNis on the order of (r/ε)N , and in particular, if ε < r, goes to infinity as N goes tothe infinity.

3.3 Bounded and Totally Bounded Sets in (`∞, dsup).

Theorem 5. In (`∞, dsup) there exist bounded sets that are not totally bounded.

Proof. It suffices to provide an example. Let A = {(1, 0, 0, . . . ), (0, 1, 0, . . . ), . . . }.A is bounded; in particular, A ⊆ N2(0). But it is not totally bounded. The dsupdistance between any two points in A is 1, so at most one point of A can be inany dsup ball of radius 1/2 (and hence diameter 1). Since there are infinitely manypoints in A, there is no finite set E such that A ⊆

⋃x∈E N1/2(x). �

4 Sequences and Completeness.

4.1 Sequences.

Let (X, d) be a metric space. An infinite sequence is a function from the naturalnumbers to X. The value that this function takes at 0 is denoted x0, the value at 1is denoted x1, and so on. The domain is called the set of indices. One can use othersets of integers (including negative numbers) as indices. In particular, I typicallyuse the natural numbers starting at 1, rather than 0. And one can work with finitesequences. All sequences here will be infinite, however. I denote a typical sequenceby (xt) = (x1, x2, . . . , xt, . . . ).

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3Some texts, including earlier versions of these notes, use {xt} = {x1, x2, . . . } as notation for asequence. This notation creates confusion with sets. For example, the sequence (1, 1, . . . ) and theset {1, 1, . . . } = {1} are very different objects.

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A sequence in X can be interpreted as a point in the countably infinite Cartesianproduct of X: both are defined as a function from the natural numbers to X. Forthis reason, I am using the same notation for both the sequence (x1, x2, . . . ) in Xand the point (x1, x2, . . . ) in X × X × . . . . In particular, a point in Rω can beinterpreted as a sequence of real numbers. Rω is, therefore, sometimes called asequence space. With this interpretation, a sequence of points in Rω is a sequenceof sequences. Since this quickly becomes confusing, I typically refer to an elementof Rω as a point rather than as a sequence.

I use subscripts to denote both elements of a sequence and, in the case of se-quences in RN or Rω, coordinates of points in the sequence. Thus, if (xt) is asequence in RN then xtn is the n coordinate of the t term of the sequence. Forexample, if the sequence is ((0, 1), (1, 5), (−1, 2), . . . ) then x3,1 is the first coordinateof the third term in the sequence, namely −1. This notation is awkward, but thealternatives are equally awkward.

The range of a sequence is the set of values that the sequence takes; formally,the range is the range of the function that defines the sequence. A sequence is saidto be in a set C ⊆ X iff the range of the sequence is a subset of C. A sequence isbounded iff its range is bounded.

4.2 Convergence.

A sequence (xt) in X is said to converge to a point x ∈ X, also written

xt → x

or limt xt = x, iff for every ε > 0 there is a T ∈ N such that for all t > T , xt ∈ Nε(x)(equivalently, d(xt, x) < ε). In words, xt → x iff all terms late in the sequence areclose to x.

Although establishing facts about convergence for important classes of sequencesis a major theme in basic analysis texts, I will not do so here. It is useful, however,to note that the sequence (1/t) in R converges to 0. To see this, note that, by theArchimedean Property of R, for any ε > 0 there is a T ∈ N such that T > 1/ε,hence 1/T < ε. This implies that 1/t < ε for all t > T , hence 1/t→ 0.

A related observation is that it suffices to check convergence only for a strictlypositive sequence (εs) that converges to 0; any such sequence will do. In view of thediscussion in the previous paragraph, a prime candidate is εs = 1/s.

Theorem 6. Let (X, d) be a metric space, let (xt) be a sequence in X and let x ∈ X.Consider any strictly positive sequence (εs) in R such that εs → 0. Then xt → x ifffor any s there is a T such that for all t > T , xt ∈ Nεs(x).

Proof. ⇒. This is immediate since any εs is simply one possible ε.⇐. Take any ε > 0. Since εs → 0, there is an s such that εs < ε. Take any T

such that for all t > T , xt ∈ Nεs(x). For any such t, since εs < ε, xt ∈ Nε(x). �

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In a metric space, a sequence cannot converge to two different points.

Theorem 7. Let (xt) be a sequence in a metric space (X, d). If xt → x∗ and xt → x̂then x∗ = x̂.

Proof. Fix any ε > 0. If xt → x∗ then there is a T ∗ such that for all t > T ∗,d(xt, x

∗) < ε/2. If xt → x̂ then there is a T̂ such that for all t > T̂ , d(xt, x̂) < ε/2.Therefore, by the triangle inequality, for any t > max{T ∗, T̂},

d(x∗, x̂) ≤ d(x∗, xt) + d(xt, x̂) < ε.

Since ε was arbitrary, this implies d(x∗, x̂) = 0, hence x∗ = x̂. �

Remark 3. For sequences in R, it is common to write xt → ∞, and to say that xtdiverges to infinity, iff for any M ∈ R there is a T such that for all t > T , xt > M .Similarly, it is common to write xt → −∞, and to say that xt diverges to negativeinfinity, iff for any M ∈ R there is a T such that for all t > T , xt < M .

Note that xt → ∞ does not mean that for any ε > 0 there is a T such that forall t > T , xt ∈ Nε(∞), since Nε(∞) has no meaning. �

Remark 4. The space X matters for the definition of convergence. Consider, forexample, the sequence of rational numbers (xt) = (1, 1.4, 1.42, 1.428, . . . ) that isgenerated by the decimal expansion of

√2. It is not hard to show that xt →

√2 if

X = R. But if X = Q then (xt) does not converge. �

4.3 Subsequences.

Given a sequence (xt), one can construct a new sequence, called a subsequence, outof the original sequence. The subsequence is of the form (xtk) = (xt1 , xt2 , . . . ) wheret1 < t2 < . . .. Even if (xt) fails to converge, some of its subsequences may converge.

Example 6. Consider the sequence (xt) = (1, 0, 1, 0, . . . ) in R. This sequence doesnot converge. I can form the subsequence (xtk) = (x1, x3, x5, . . . ) = (1, 1, 1, . . . ).This subsequence converges to 1. There are, of course, other subsequences. Forexample, consider (x10, x11, x20, x21, . . . ) = (0, 1, 0, 1, . . . ). This does not converge.�

Theorem 8. Let (X, d) be a metric space. A sequence (xt) in X converges to apoint x∗ ∈ X iff every subsequence of (xt) converges to x∗.

Proof. Exercise. �

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4.4 Cauchy sequences and completeness.

Definition 2. Let (X, d) be a metric space. A sequence (xt) in X is Cauchy iff forany ε > 0 there is a T such that for any s, t > T , d(xs, xt) < ε.

In words, a sequence is Cauchy iff all terms late in the sequence are close to eachother. The next theorem establishes that all convergent sequences are Cauchy.

Theorem 9. Let (X, d) be a metric space. Let (xt) be a sequence in X. If (xt) isconvergent then it is Cauchy.

Proof. Suppose xt → x. Consider any ε > 0. Choose T such that for all t > T ,d(xt, x) < ε/2. Then, by the triangle inequality, for any s, t > T , d(xs, xt) ≤d(xs, x) + d(x, xt) < ε. �

If the converse is always true, if Cauchy sequences always converge, then the setis called complete.

Definition 3. Let (X, d) be a metric space. X is complete iff for any sequence (xt)in X, if (xt) is Cauchy then there is a point x ∈ X such that xt → x.

If (X, d) is a metric space and A ⊆ X then the statement that A is completemeans that A is complete when viewed as a metric subspace of X. I will refer to Aas either a complete space or a complete set.

Later in the course, I establish that R is complete. In contrast, the space ofrational numbers, Q, is not complete.

Example 7. Consider the Cauchy sequence (3, 3.1, 3.14, 3.141, . . . ). Considered asa sequence in R, this sequence converges to π. But π 6∈ Q. As a sequence in themetric space Q, therefore, this Cauchy sequence does not converge. �

To sum up, informally, a Cauchy sequence is a sequence that looks like it con-verges. If the sequence lies in a complete space then a Cauchy sequence actuallydoes converge. But a Cauchy sequence in a space such as Q that is not completeneed not converge.

For later use, I record the following additional facts about Cauchy sequences.

Theorem 10. Let (X, d) be a metric space. Let (xt) be a Cauchy sequence in X.Then (xt) is bounded.

Proof. Since (xt) is Cauchy, there is a T such that for all t, s > T , d(xs, xt) < 1.In particular, for all t > T , d(xt, xT+1) < 1. Let

r = max{d(x1, xT+1), . . . , d(xT , xT+1)}+ 1

Then the range of (xt) lies in Nr(xT+1). �

Note that, combined with Theorem 9, this implies that any convergent sequenceis bounded.

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Theorem 11. Let (X, d) be a metric space. Let (xt) be a Cauchy sequence in X.If (xt) has a convergent subsequence then it is convergent.

Proof. Suppose xtk → x and choose any ε > 0. Since xtk → x, there is a K suchthat if k > K then xtk ∈ Nε/2(x). Since (xt) is Cauchy there is a T such that ifs, t > T , d(xs, xt) < ε/2. Then, by the triangle inequality, for any t > T and anytk > max{T, tK}

d(xt, x) ≤ d(xt, xtk) + d(xtk , x) < ε.

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Remark 5. Completeness as used here refers to Cauchy completeness. There areother forms of completeness used in mathematics, notably Dedekind completeness.See Propp (2014). �

5 Open and Closed Sets.

5.1 Open sets.

Definition 4. Let (X, d) be a metric space and let A ⊆ X. A point x ∈ A is interiorto A iff it is contained in a ball inside A: there is an ε > 0 such that Nε(x) ⊆ A.

Theorem 12. Let (X, d) be a metric space and consider any O ⊆ X. The followingare equivalent.

1. O is either empty or the union of balls.

2. Every point x ∈ O is interior to O.

Proof. If O is empty then the equivalence is immediate. Therefore, assume that Ois not empty.⇒. Consider any x ∈ O. By property 1, there is an a ∈ O and an εa > 0

such that x ∈ Nεa(a) ⊆ O. I need to show that there is an ε > 0 such thatNε(x) ⊆ O. Let ε = εa − d(a, x). Consider any b ∈ Nε(x). By the triangleinequality, d(b, a) ≤ d(b, x) + d(x, a) < εa. Hence b ∈ Nεa(a), which implies thatb ∈ O, which implies that Nε(x) ⊆ O, as was to be shown.⇐ . For each x ∈ O, choose εx > 0 such that Nεx(x) ⊆ O. Then

O =⋃x∈O

Nεx(x),

as was to be shown. �

Any set that satisfies either of the properties in Theorem 12, and hence both ofthese properties, is called open. Formally, I define openness using property 1.

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Definition 5. Let (X, d) be a metric space. A set O ⊆ X is open iff it is eitherempty or the union of balls.

Example 8. In R, any interval (a, b) is a ball and therefore is open: set ε = (b−a)/2and x = (a + b)/2; then (a, b) = Nε(x). Alternatively, one can appeal to Theorem12 and check that every x ∈ (a, b) is interior: set ε = min{x − a, b − x} and verifythat Nε(x) ⊆ (a, b). �

Whether a set A ⊆ X is open is not intrinsic to A but, rather, depends on thecomplement, Ac = X \A, which depends on X. If (X, d) is a metric space, X̂ ⊆ X,and A ⊆ X̂, then it is possible for A to be open in the metric space defined by X̂(with the metric inherited from X) even if it is not open in X.

Example 9. Let A = [0, 1).

• If X = R then A is not open. In particular, 0 is not interior to A for X = Rsince, for any ε > 0, Nε(0) = (−ε, ε) 6⊆ [0, 1).

• If X = [0, 1), hence X = A, then A = X is open. In particular, 0 is nowinterior, since, in the X = [0, 1) metric space, for any ε ∈ (0, 1), Nε(0) = {x ∈[0, 1) : d(x, 0) < ε} = [0, ε) ⊆ [0, 1).

�

Example 10. Let A = Q.

• If X = R then A is not open. In particular, it is not hard to show that forany x ∈ Q and any ε > 0, Nε(x) must contain irrational numbers and hencecannot be a subset of Q.

• If X = Q (hence X = A), then A is open. In particular, any x ∈ Q is nowinterior since, in the X = Q metric space, for any ε > 0, Nε(x) = {q ∈ Q :d(q, x) < ε} = (x− ε, x+ ε) ∩Q ⊆ Q.

�

The next result establishes the fundamental properties of open sets.

Theorem 13. Let (X, d) be a metric space.

1. Both X and ∅ are open.

2. For any two open sets O1 and O2, O1 ∩ O2 is open. By induction, finiteintersections of open sets are open.

3. For any set O of open sets, ⋃O∈O

O

is open. Thus, arbitrary unions of open sets are open.

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Proof.

1. X is open since it is the union of all balls. ∅ is open by definition.

2. Let O1 and O2 be open. If O1 ∩ O2 = ∅ then O1 ∩ O2 is open by definition.Otherwise, let x ∈ O1 ∩ O2. Since x ∈ O1 and O1 is open there is an ε1 > 0such that Nε1(x) ⊆ O1 (Theorem 12). Similarly, there is an ε2 > 0 such thatNε2(x) ⊆ O2. Set ε = min{ε1, ε2}. Then Nε(x) ⊆ O1 ∩O2.

3. If the O are open then⋃O∈O O is a union of unions of balls, and is therefore

open by definition.

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The next example shows that arbitrary intersections of open sets need not beopen.

Example 11. Suppose O consists of intervals of the form (0, 1+1/t) for t ∈ {1, 2, . . . ).Then ⋂

O∈OO = (0, 1],

which is not open. �

A metric space is a special case of a more general type of space called a topologicalspace. A topological space specifies a set of points X and a set of open sets τ (τ for‘topology’), with the requirement that the sets in τ must satisfy the conditions inTheorem 13. The twist is that the sets in τ don’t have to be derived from a metricand in fact there are topological spaces that do not correspond to any metric space.For example, let X = R and let τ = {X, ∅}: the only open sets are the space itselfand the empty set. There is no metric on R that that can generate this τ . Themain result on existence of optima generalizes to arbitrary topological spaces. Ofthe other results, some generalize and some do not.

5.2 Open sets and sequences.

Theorem 14. Let (X, d) be a metric space, let (xt) be a sequence in X and letx∗ ∈ X. xt → x∗ iff for any open set O containing x∗, there is a T ∈ N such thatfor all t > T , xt ∈ O.

Proof. ⇒. Let O be an open set containing x∗. Since O is open, x∗ is interior, andhence there is an ε > 0 such that Nε(x

∗) ⊆ O. Since xt → x∗, there is a T such thatfor all t > T , xt ∈ Nε(x

∗) ⊆ O, which is what I needed to show.⇐. The claim is immediate since Nε(x

∗) is open for any ε > 0. �

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5.3 Limit points.

Theorem 15. Let (X, d) be a metric space, let A ⊆ X, and let x ∈ X. The followingare equivalent.

1. For any ε > 0 there is an a ∈ A ∩Nε(x) such that a 6= x.

2. For any ε > 0, A ∩Nε(x) is infinite.

3. There is a sequence (at) in A such that at → x and for all s, t, as 6= at andat 6= x.

Proof.

1. (1) ⇒ (2). By contraposition. Suppose that there is an ε > 0 such thatA ∩ Nε(x) is finite. If A ∩ Nε(x) \ {x} is either empty, then (1) is violated.Therefore, suppose that A ∩ Nε(x) \ {x} equals a non-empty finite set B ={a1, . . . , aK}. Let δ = minak∈B d(ak, x). Then δ > 0, since B is finite andak 6= x for all k. But then A ∩Nδ(x) \ {x} is empty and (1) is violated.

2. (2) ⇒ (3). Choose a1 to be any point in A ∩ N1(x) \ {x}. Since A ∩ N1(x)is infinite, this is must be possible. In general, for any t, choose at to be anypoint in A∩N1/t(x) \ {x} that is not equal to any as, s < t. Since A∩N1/t(x)is infinite, this must be possible. The sequence (at) has the desired properties.

3. (3) ⇒ (1). From (3), there is a sequence (at) in A such that for any ε > 0there is a T such that for all t > T , at ∈ A∩Nε(x)\{x}. Hence (1) is satisfied.

�

A point x ∈ X is said to be a limit point of a set A ⊆ X iff it satisfies any of theproperties in Theorem 15, and hence all of the properties in Theorem 15. Formally,I define limit point using property 1.

Definition 6. Let (X, d) be a metric space and let A ⊆ X.

1. x ∈ X is a limit point of A iff for any ε > 0 there is an a ∈ A ∩Nε(x) suchthat a 6= x.

2. x ∈ X is an isolated point of A iff x ∈ A and x is not a limit point of A.

Remark 6. Note that by Theorem 15 property 2, a set has to be infinite in order tohave any limit points. Thus, for example, the set A = {0} has no limit points. �

Remember that limits are for sequences. Limit points are for sets. The conceptsare related but distinct. Theorem 15 shows that a limit point of a set is the limit ofa sequence drawn from that set. The next result establishes a partial converse.

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Theorem 16. Let (X, d) be a metric space. Let (xt) be a sequence in X and let Abe the range of (xt). If xt → x, then either x is a limit point of A or A is finite andx ∈ A.

Proof. Suppose that xt → x and suppose that x is not a limit point of A. Thenthere is an ε > 0 such that Nε(x) ∩ A is either empty or contains only x. Sincext → x, this implies that there is a T such that for all t > T , xt = x, in which caseA is finite and x ∈ A. �

Here are some examples that may help clarify some of these concepts.

Example 12. 0 is a limit point of both [0, 1] and (0, 1]. �

Example 13. Let (xt) be the sequence (1, 1/2, 1/3, . . . ) and let A be the range ofthis sequence. Then xt → 0. 0 is the only limit point of A. Every point of A isisolated.

Example 14. Let (xt) be the sequence (0, 0, 0, . . . ) and let A be the range of thissequence: A = {0}. Then xt → 0. A has no limit points. �

Example 15. A = [0, 1]. Then 0 is a limit point of A, but A, which is uncountable,is not the range of a sequence (which must be countable). �

Remark 7. If (xt) is a sequence in X then x ∈ X is an accumulation point or clusterpoint of (xt) iff for any ε > 0 there are infinitely many t such that xt ∈ Nε(x). It isalmost immediate that x is an accumulation point of (xt) iff there is a subsequence of(xt) converging to x: “accumulation point” is just another way to say “subsequentiallimit.”

Let A be the range of (xt). If x is a limit point of A then x is an accumulationpoint of (xt) and, moreover, there is a subsequence of (xt) that converges to x andin which all terms have different values.

On the other hand, an accumulation point of (xt) need not be a limit point ofA. For example, if (xt) = (1, 1, 1, . . . ) then A = {1}, which has no limit points, butx = 1 is an accumulation point of (xt). �

5.4 Closed sets.

Theorem 17. Let (X, d) be a metric space and consider any C ⊆ X. The followingare equivalent.

1. For any sequence (xt) in C, if xt → x then x ∈ C.

2. For any x ∈ X, if x is a limit point of C then x ∈ C.

3. Cc is open.

Proof.

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1. (1) ⇒ (2). Let x be a limit point of C. By Theorem 15, there is a sequence(xt) in C such that xt → x. By (1), x ∈ C.

2. (2) ⇒ (3). By contraposition. Suppose that (3) fails. Then Cc is not open.Then there exists an x ∈ Cc such that x is not interior to Cc. Hence for anyε > 0, Nε(x) 6⊆ Cc, hence Nε(x) ∩ C 6= ∅. Take any a ∈ Nε(x) ∩ C. Thena 6= x since x ∈ Cc. Since ε was arbitrary, this shows that x is a limit pointof C. But since x /∈ C, (2) fails.

3. (3)⇒ (1). By contraposition. Suppose that (1) fails. Then there is a sequence(xt) in C and an x ∈ X such that xt → x but x 6∈ C. Since x 6∈ C, x ∈ Cc.Take any ε > 0. Since xt → x, there is an xt ∈ Nε(x). Since xt ∈ C, hencext 6∈ Cc, Nε(x) 6⊆ Cc. Since ε was arbitrary, this shows that x is not interiorto CC , hence Cc is not open, hence (3) fails.

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Any set satisfying any of the properties in Theorem 17, hence satisfying all ofthem, is called closed. Formally, I define closed using property 3.

Definition 7. Let (X, d) be a metric space. A set C ⊆ X is closed iff its comple-ment, Cc, is open.

The next result is the analog of Theorem 13.

Theorem 18. Let (X, d) be a metric space.

1. Both X and ∅ are closed.

2. For any two closed sets C1 and C2, C1 ∪ C2 is closed. By induction, finiteunions of closed sets are closed.

3. For any set C of closed sets ⋂C∈C

C

is closed. Thus, arbitrary intersections of closed sets are closed.

Proof. Immediate from Theorem 13 and DeMorgan’s laws (see the Set Theorynotes). �

The next example shows that arbitrary unions of closed sets need not be closed.

Example 16. Suppose O consists of intervals of the form [0, 1−1/t] for t ∈ {1, 2, . . . ).Then ⋃

O∈OO = [0, 1),

which is not closed. �

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The terms “open” and “closed” can be misleading in the following respect. InEnglish, a door, say, can be either open, closed, or partly open, but not both openand closed. This is not true for open and closed for sets. We have already seen that,in an metric space (X, d), both X and ∅ are both open and closed.

A related point is that in RN , RN and ∅ are the only sets that are both closedand open. But there are other spaces in which this is not true. For example, considerthe space X = R\{0} with the standard Euclidean metric. In this space, (−∞, 0) isboth open and closed, as is (0,∞). This X violates a property called connectedness.It is, in fact, the case that for any metric space X, X is connected iff X and ∅ arethe only sets that are both open and closed. See the notes on Connected Spaces.

Another subtlety is that the property of a set being closed is not intrinsic to theset but also depends on the ambient space X, as illustrated by the next example;see also the related discussion of open sets and Example 9.

Example 17. Let C = [0, 1).

• If X = R then C is not closed, since 1 is a limit point of C but 1 /∈ C.

• But if X = [0, 1), so that C = X, then C is closed (remember that X is alwaysclosed). In effect, if X = [0, 1) then the number 1 does not exist.

�

In contrast to the property of being closed, the property of being complete isintrinsic to the set; it does not depend on the ambient space. The next resultestablishes that a complete set is always closed, regardless of the ambient space.The converse is not true, as illustrated by the previous example: whether C = [0, 1)is closed depends on whether the space is X = [0, 1) or X = R, but either way, C isnot complete.

Theorem 19. Let (X, d) be a metric space and let C ⊆ X.

1. If C is complete then C is closed.

2. If X is complete and C is closed then C is complete.

Proof.

1. Let (xt) be a sequence in C and suppose that there is an x ∈ X such thatxt → x. Since xt → x, (xt) is Cauchy (Theorem 9). Since C is complete, thereis an x∗ ∈ C such that xt → x∗. Since also xt → x, x = x∗ (Theorem 7), andhence x ∈ C, hence C is closed.

2. Take any Cauchy sequence (xt) in C. Since C ⊆ X, (xt) is a Cauchy sequencein X. Since X is complete, there is an x ∈ X such that xt → x. Since C isclosed, x ∈ C. Hence C is complete.

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�

In practice, we will be working mainly with spaces X that are complete, inwhich case, by the second part of Theorem 19, the statement that C ⊆ X is closedis equivalent to the statement that C is complete; but to take advantage of thisequivalence, we may first have to prove that X is complete.

Given a set A, the closure of A is the smallest closed set containing A.

Definition 8. Given a set A ⊆ X, A, called the closure of A, is the intersection ofall closed sets containing A.

This definition is not vacuous since X itself is a closed set containing A. Asan intersection of closed sets, A is closed. It is immediate from the definition thatA ⊆ A.

Given a set A, let A′ denote the set of limit points of A.

Theorem 20. Let (X, d) be a metric space. For any set A ⊆ X,

A = A ∪A′.

Proof. Since A is closed and A ⊆ A, it follows from Theorem 17 that A ∪A′ ⊆ A.Since A ⊆ A ∪ A′, it suffices to prove that A ∪ A′ is closed, since then A ⊆

A ∪A′ = A ∪A′ and hence A = A ∪A′.To show that A∪A′ is closed, I argue by contraposition. Consider any x /∈ A∪A′.

Therefore, x /∈ A and x is not a limit point of A. Hence there is an ε > 0 such thatNε(x) ⊆ Ac. Consider any point b ∈ Nε(x). Then b 6∈ A. Also, since Nε(x) is open,b is interior to Nε(x), hence there is an εb such that Nεb(b) ⊆ Nε(x) ⊆ Ac. Thisimplies that b /∈ A′. Therefore, b /∈ A ∪ A′. Since this holds for any b ∈ Nε(x), x isnot a limit point of A ∪ A′. In summary, if x is not in A ∪ A′ then it is not a limitpoint of A ∪ A′. By contraposition, of x is a limit point of A ∪ A′ then x ∈ A ∪ A′,as was to be shown. �

In R, finite unions of closed intervals are closed, but closed sets in R can be muchmore complicated than finite unions of closed intervals.

Example 18. Let E0 = [0, 1]\(1/3, 2/3) = [0, 1/3]∪[2/3, 1], let E1 = E0\((1/9, 2/9)∪(7/9, 8/9)) = [0, 1/9] ∪ [2/9, 1/3] ∪ [2/3, 7/9] ∪ [8/9, 1] and so on. In words, ateach stage, remove the (open) middle third from each remaining subinterval. LetA =

⋂Et. A is called the Cantor set.

A is closed, since it is an intersection of closed sets. In fact, one can show that Ais perfect, meaning that A = A′. And A is uncountable.4 But Ac is open (since A is

4This is true in general of perfect sets in R. But it can be seen as follows. Write the elements of[0,1] as ternary expansions (i.e., in the form x = x1/3+x2/9+x3/27+. . . ) and then note that x ∈ Aiff it has a ternary expansion with no 1s in it. Thus x ∈ A iff it has a ternary expansion that is astring of 0s and 2s. Writing a 1 in place of a 2 puts the set of such strings into 1-1 correspondencewith the set of strings of 0s and 1s, which can be identified with the set of binary expansions ofelements of [0, 1].

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closed) and one can show that Ac is dense in [0, 1], meaning that Ac = [0, 1]. So Ais large in one sense (it is uncountable) but small in another (it is the complementof a set that is open and dense). �

Intuitively, the closed ε ball, Nε(x), should equal {a ∈ X : d(a, x) ≤ ε}. Thisturns out to be true in normed vector spaces but not in general.

Theorem 21. Let (X, d) be a metric space. For any x ∈ X and any ε > 0,Nε(x) ⊆ {a ∈ X : d(a, x) ≤ ε}.

Proof. By Theorem 20, since Nε(x) ⊆ {a ∈ X : d(a, x) ≤ ε}, it remains to show ifa is a limit point of Nε(x), then d(a, x) ≤ ε. Equivalently, I need to argue that if(at) is a sequence in Nε(x) and at → a then d(a, x) ≤ ε. By the triangle inequality,since at ∈ Nε(x), d(a, x) ≤ d(a, at) + d(at, x) < d(a, at) + ε. Since at → a, d(a, at)can be made arbitrarily small, which implies d(a, x) ≤ ε. �

A corollary of Theorem 21 is that the closure of a bounded set is bounded.

Theorem 22. Let (X, d) be a metric space. If B ⊆ X is bounded then B is alsobounded.

Proof. Since B is bounded, there is an x ∈ X and an r > 0 such that B ⊆ Nr(x),hence B ⊆ Nr(x) ⊆ {a ∈ X : d(a, x) ≤ r} ⊆ Nr+1(x). �

In metric spaces that are not normed vector spaces, the set inclusion Nε(x) ⊆{a ∈ X : d(a, x) ≤ ε} can be strict.

Example 19. Consider (X, dd) where dd, the discrete metric, was defined in Section1.3.3. For any x ∈ X, N1(x) = {x}, N1(x) = {x}, but {a ∈ X : d(a, x) ≤ 1} = X.�

Theorem 23. Let X be a vector space with norm ‖ · ‖ and associated metric d. Forany x ∈ X and any ε > 0, Nε(x) = {a ∈ X : d(a, x) ≤ ε}.

Proof. For ease of notation, take x = 0. Take any ε > 0. By Theorem 21, it sufficesto show that if ‖a‖ = ε, then a is a limit point of Nε(0). Since a /∈ Nε(0), this meansshowing that for any δ > 0, Nε(0) ∩Nδ(a) 6= ∅. It suffices to restrict attention to δthat is small, specifically δ < ε. Take any γ ∈ (1− δ/ε, 1). Since γ < 1, γa ∈ Nε(0).And d(a, γa) = ‖a − γa‖ = (1 − γ)‖a‖ = (1 − γ)ε < δ (since γ > 1 − δ/ε), henceγa ∈ Nδ(a). �

Finally, I note that in R, the inf and sup of a bounded set that is closed mustbe in the set.

Theorem 24. Let C ⊆ R be closed. If C is bounded above, then supC ∈ C.Similarly, if C is bounded below, then inf C ∈ C.

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Proof. If C is bounded above then by the LUB property, C has a least upperbound, say b ∈ R. Since b is the least upper bound, for any t ∈ {1, 2, . . . }, b − 1/tis not an upper bound, and hence there is an xt ∈ C such that xt > b− 1/t. Sinceb is an upper bound, xt ≤ b. For the sequence (xt) thus formed, for any t ≥ T ,xt ∈ N1/T (b). Thus xt → b. Since C is closed, b ∈ C.

The proof for C bounded below is almost identical. �

6 Strongly Equivalent Metrics.

Definition 9. Let X be a set. Two metrics d1 and d2 on X are strongly equivalentiff there exist real numbers 0 < a ≤ b such that for any x, y ∈ X,

ad2(x, y) ≤ d1(x, y) ≤ bd2(x, y). (1)

Recall that the notes on Vector Spaces and Norms introduced the very similarconcept of equivalent norms.

Theorem 25. Let X be a vector space, let f1 and f2 be norms on X, and let d1and d2 be the associated induced metrics. Then f1 and f2 are equivalent iff d1 andd2 are strongly equivalent.

Proof. Almost immediate from the definitions. �

One implication is that in any finite dimensional vector space, all norm-basedmetrics are strongly equivalent, since (as discussed in the notes on Vector Spacesand Norms and proved in the notes on Existence of Optima) all norms on a finitedimensional vector space are equivalent.

The main result of this section, Theorem 27, is that if two metrics are stronglyequivalent then they have the same bounded sets, totally bounded sets, convergentsequences, Cauchy sequences, open sets, closed sets, and complete sets.

I begin by noting, in Theorem 26, that if two metrics on X are strongly equiva-lent, then, around any given point x∗ ∈ X, they generate open balls that are nested,like matryoshka dolls. (I made a similar observation in the notes on Vector Spaces.)

As notation, if d1 and d2 are two metrics on X, then, for any ε > 0 and anyx∗ ∈ X, let Nε(x

∗)|1 denote the open ε ball centered on x∗ under the d1 metric, andlet Nε(x

∗)|2 denote the open ε ball centered on x∗ under the d2 metric.

Theorem 26. Let d1 and d2 be strongly equivalent metrics on X. Then for anyε1 > 0 there are ε̂2 ≥ ε2 > 0 such that for any x∗,

Nε2(x∗)|2 ⊆ Nε1(x∗)|1 ⊆ Nε̂2(x∗)|2. (2)

Conversely, for any ε2 > 0 there are ε̂1 ≥ ε1 > 0 such that for any x∗,

Nε1(x∗)|1 ⊆ Nε2(x∗)|2 ⊆ Nε̂1(x∗)|1. (3)

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Proof. Since d1 and d2 are strongly equivalent, there exist 0 < a ≤ b such that forany x, y ∈ X, Inequality 1 holds.

1. Subset Relation 2. Fix ε1 > 0 and let ε2 = ε1/b and ε̂2 = ε1/a; since b ≥ a > 0,these are well defined with ε̂2 ≥ ε2. Take any x∗.

• To see that Nε2(x∗)|2 ⊆ Nε1(x∗)|1, consider any x ∈ Nε2(x∗)|2. Thend2(x, x

∗) < ε2 = ε1/b, which implies bd2(x, x∗) < ε1. By Inequality 1,

this implies d1(x, x∗) < ε1, which implies x ∈ Nε(x

∗)|1, which provesNε2(x∗)|2 ⊆ Nε1(x∗)|1.• To see that Nε1(x∗)|1 ⊆ Nε̂2(x∗)|2, consider any x ∈ Nε1(x∗)|1. Thend1(x, x

∗) < ε1. By inequality 1, this implies ad2(x, x∗) < ε1, which

implies d2(x, x∗) < ε1/a = ε̂2. Hence x ∈ Nε̂2(x∗)|2, which proves

Nε1(x∗)|1 ⊆ Nε̂2(x∗)|2.

2. Subset Relation 3. Fix ε2 > 0 and set ε1 = aε2 and ε̂1 = bε2. The rest of theargument is so similar that I omit it.

�

Theorem 27. Let X be a set and let d1 and d2 be strongly equivalent metrics onX.

1. For any A ⊆ X, A is bounded under d1 iff it is bounded under d2.

2. For any A ⊆ X, A is totally bounded under d1 iff it is totally bounded underd2.

3. For any sequence (xt) and point x∗ in X, (xt) converges to x∗ under d1 iff itconverges to x∗ under d2.

4. For any sequence (xt) in X, (xt) is Cauchy under d1 iff it is Cauchy underd2.

5. For any A ⊆ X, A is open under d1 iff it is open under d2.

6. For any A ⊆ X, A is closed under d1 iff it is closed under d2.

7. For any A ⊆ X, A is complete under d1 iff it is complete under d2.

Proof. For each iff claim, the iff proof is the same in both directions apart fromlabeling and so I provide only one direction.

1. Boundedness. If A is d1 bounded then there is an r1 > 0 and an x∗ ∈ X suchthat A ⊆ Nr1(x∗)|1. By Theorem 26 (Subset Relation 2), there is an r2 > 0such that Nr1(x∗)|1 ⊆ Nr2(x∗)|2, which shows that A is d2 bounded.

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2. Total Boundedness. Suppose that A is d1 totally bounded. Take any ε2 > 0and let ε1 > 0 be as in Theorem 26 (Subset Relation 3). Since A is d1 totallybounded there is a finite set E such that A ⊆

⋃x∈E Nε1(x)|1. For each x ∈ E,

we have Nε1(x)|1 ⊆ Nε2(x)|2 and so A ⊆⋃x∈E Nε2(x)|2, which implies that A

is d2 totally bounded.

3. Convergence. Suppose that (xt) d1 converges to x∗. Take any ε2 > 0 and letε1 be as in Theorem 26 (Subset Relation 3). Since (xt) dt converges to x∗,there is a T such that for all t > 0T , d1(xt, x

∗) < ε1, hence d2(xt, x∗) < ε2,

which implies that (xt) d2 converges to x∗.

4. Cauchy. Suppose that (xt) is d1 Cauchy. Take any ε2 > 0 and let ε1 be as inTheorem 26 (Subset Relation 3). Since (xt) is d1 Cauchy, there is a T suchthat for all s, t > T , d1(xs, xt) < ε1, which implies d2(xs, xt) < ε2, whichimplies that (xt) is d2 Cauchy.

5. Open. Let A be d1 open. Consider any x ∈ A. Since A is d1 open, x is d1interior to A: there exists an ε1 > 0 such that Nε1(x)|1 ⊆ A. Let ε2 > 0 beas in Theorem 26 (Subset Relation 2) with Nε2(x)|2 ⊆ Nε1(x)|1. Then x is d2interior to A and hence A is open.

6. Closed. This follows from the previous result for open sets and the definitionof a closed set as the complement of an open set.

7. Complete. Let A be d1 complete. If (xt) is d2 Cauchy then (by the previousresult for Cauchy sequences) (xt) is also d1 Cauchy, which implies, since A isd1 complete, that there is an x∗ ∈ A such that (xt) d1 converges to x∗, whichimplies (by the previous result for convergence) that (xt) d2 converges to x∗.

�

The proof that, in RN , bounded sets are totally bounded in RN follows quicklyfrom Theorem 27.

Proof of Theorem 4. In the notes on Vectors Spaces and Norms, I showed that theEuclidean and max norms on RN are equivalent, which implies that their inducedmetrics are strongly equivalent. In view of Theorem 27, it therefore suffices to provethat, under dmax, bounded sets in RN are totally bounded.

Suppose that A ⊆ RN is bounded under dmax. Then there is an r such that Ais contained in an open r ball, and is therefore contained in a closed r-ball. Underdmax, a closed r ball is an N -dimensional cube with edges of length 2r. Fix anyε > 0 and take any integer T > r/ε. Divide the cube into TN subcubes, each withedges of length 2r/T . Since T > r/ε, each subcube is contained in an open ε ball(which, under dmax, is an open cube). Hence A is covered by a finite (TN ) number

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of open ε balls. A is, therefore, totally bounded. �

In infinite dimensional vector spaces, different norms may not be equivalent and,as a consequence, the induced metrics may have different convergence properties.The following example illustrates what can go wrong.

Example 20. Let X = `∞ and consider the sup and p∗ norms, where the latter (firstintroduced in the notes on Vector Spaces and Norms) is given by

‖x‖p∗ = supn

|xn|n.

Let dsup and dp∗ be the associated metrics. These norms are not equivalent (asalready discussed in the notes on Vector Spaces and Norms), and the associatedmetrics are not strongly equivalent.

Consider the sequence (xt) with terms x1 = (1, 0, 0, . . . ), x2 = (0, 1, 0, 0, . . . ), andso on. Then this sequence converges to the origin under dp∗ but does not converge(to anything) under dsup. �

References

Propp, James. 2014. “Real Analysis in Reverse.” The American MathematicalMonthly 120(5):392–408.

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