3.3. LIMIT POINTS 95

3.3 Limit Points

3.3.1 Main Denitions

Intuitively speaking, a limit point of a set S in a space X is a point of X whichcan be approximated by points of S other than x as well as one pleases.The notion of limit point is an extension of the notion of being "close" to a

set in the sense that it tries to measure how crowded the set is. To be a limitpoint of a set, a point must be surrounded by an innite number of points ofthe set.We now give a precise mathematical denition. In what follows, R is the

reference space, that is all the sets are subsets of R.

Denition 263 (Limit point) Let S R, and let x 2 R.1. x is a limit point or an accumulation point or a cluster point of Sif 8 > 0, (x ; x+ ) \ S n fxg 6= ?.

2. The set of limit points of a set S is denoted L (S)

Remark 264 Let us remark the following:

1. In the above denition, we can replace (x ; x+ ) by a neighborhood ofx. Therefore, x is a limit point of S if any neighborhood of x containspoints of S other than x.

2. Being a limit point of a set S is a stronger condition than being close to aset S. It requires any neighborhood of the limit point x to contain pointsof S other than x.

3. The above two remarks should make it clear that L (S) S. We can alsoprove it rigorously. If x 2 L (S) then 9 > 0 : (x ; x+ )\S nfxg 6= ?.Let y 2 (x ; x+ )\ S n fxg. Then, y 2 (x ; x+ ) and y 2 S n fxg.So, y 2 S thus (x ; x+ ) \ S 6= ?. Hence x 2 S.

4. However, S is not always a subset of L (S).

Let us rst look at easy examples to understand what a limit point is andwhat the set of limit points of a given set might look like.

Example 265 Let S = (a; b) and x 2 (a; b). Then x is a limit point of (a; b).Let > 0 and consider (x ; x+ ). This interval will contain points of (a; b)other than x, innitely many points in fact.

Example 266 a and b are limit points of (a; b). Given > 0, the interval(a ; a+ ) contains innitely many points of (a; b) n fag thus showing a is alimit point of (a; b). The same is true for b.

Example 267 Let S = [a; b] and x 2 [a; b]. Then x is a limit point of [a; b].This is true for the same reasons as above.

96 CHAPTER 3. TOPOLOGY OF THE REAL LINE

At this point you may think that there is no dierence between a limit pointand a point close to a set. Consider the next example.

Example 268 Let S = (0; 1) [ f2g. 2 is close to S. For any > 0, f2g (2 ; 2 + )\S so that (2 ; 2 + )\S 6= ?. But 2 is not a limit point of S.(2 :1; 2 + :1) \ S n f2g = ?.

Remark 269 You can think of a limit point as a point close to a set but alsosurrounded by many (innitely many) points of the set.

We look at more examples.

Example 270 L ((a; b)) = [a; b]. From the rst two examples of this section,we know that L ((a; b)) [a; b]. What is left to show is that if x =2 [a; b] then xis not a limit point of (a; b). Let x 2 R such that x =2 (a; b). Assume x > b (thecase a < a is similar and left for the reader to check). Let < jx bj. Then(x ; x+ ) \ (a; b) n fxg = ?.

Example 271 L (Z) = ?. Let x 2 R. For x to be a limit point, we must havethat for any > 0 the interval (x ; x+ ) would have to contain points ofZ other than x. Since for every real number x there exists an integer n suchthat n 1 x < n, if < min (jx n+ 1j ; jx nj), the interval (x ; x+ )contains no integer therefore (x ; x+ ) \ Z n fxg = ?. So, no real numbercan be a limit point of Z.

Example 272 L (Q) = R. For any real number and for any > 0, the in-terval (x ; x+ ) contains innitely many points of Q other than x. Thus,(x ; x+ ) \Q n fxg 6= ?. So every real number is a limit point of Q.

Example 273 If S = (0; 1)[f2g then S = [0; 1][f2g but L (S) = [0; 1]. Usingarguments similar to the arguments used in the previous examples, one can provethat every element in [a; b] is a limit point of S and every element outside of[a; b] is not. You will notice that 2 2 S but 2 =2 L (S). Also, 0 and 1 are in L (S)but not in S. So, in the most general case, one cannot say anything regardingwhether S L (S) or L (S) S.

Example 274 If S =

1

n: n 2 Z+

then L (S) = f0g (see problems).

3.3.2 Properties

Theorem 275 Let x 2 R and S R.

1. If x has a neighborhood which only contains nitely many members of Sthen x cannot be a limit point of S.

2. If x is a limit point of S then any neighborhood of x contains innitelymany members of S.

3.3. LIMIT POINTS 97

Proof. We prove each part separately.

Part 2: This part follow immediately from part 1.

Part 1: Let U be a neighborhood of x which contains only a nite number ofpoints of S that is U \ S is nite. Then, U \ S n fxg is also nite.Suppose U \ S n fxg = fy1; y2; :::; yng. We show there exists > 0such that (x ; x+ ) \ U does not contain any member of S n fxg.Since both (x ; x+ ) and U are neighborhood of x, so is their inter-section. This will prove there is a neighborhood of x containing no el-ement of S n fxg hence proving x is not a limit point of S. Let =min fjx y1j ; jx y2j ; :::; jx ynjg. Since x is not equal to yi, > 0.Then (x ; x+ )\U Contains no points of S other than x thus provingour claim.

Corollary 276 No nite set can have a limit point.Proof. Follows immediately from the theorem.

The next theorem is very important. It helps understand the relationshipbetween the set of limit points of a set and the closure of a set.

Theorem 277 Let S RS = S [ L (S)

Proof. We show inclusion both ways.

1. S [ L (S) SWe already know that S S. We now show that L (S) S. This willimply the result. Suppose that x 2 L (S). Then 8 > 0, (x ; x+ ) \S n fxg 6= ?. It follows that (x ; x+ ) \ S 6= ?, since S n fxg S.Therefore, x is in the closure of S.

2. S S [ L (S)Let x 2 S. We need to prove that x 2 S [ L (S). Either x 2 S or x =2 S.If x 2 S then x 2 S [ L (S). If x =2 S then x is close to S. Therefore,8 > 0, (x ; x+ ) \ S 6= ?. Since x =2 S, S = S n fxg, therefore(x ; x+ ) \ S n fxg 6= ?. It follows that x 2 L (S) and thereforex 2 S [ L (S). So we see that in all cases, if we assume that x 2 S thenwe must have x 2 S [ L (S).

3.3.3 Important Facts to Remember

Denitions properties and theorems in this section. L (A) [ L (B) = L (A [B) (see problems)

98 CHAPTER 3. TOPOLOGY OF THE REAL LINE

S = S [ L (S) L (S) S L (S) is closed (see problems). If U is open then L (U) = U (see problems).

3.3.4 Exercises

1. Prove that if S =

1

n: n 2 Z+

then L (S) = f0g

2. In each situation below, give an example of a set which satises the givencondition.

(a) An innite set with no limit point.

(b) A bounded set with no limit point.

(c) An unbounded set with no limit point.

(d) An unbounded set with exactly one limit point.

(e) An unbounded set with exactly two limit points.

3. Prove that if A and B are subsets of R and A B then L (A) L (B).4. Prove that if A and B are subsets of R then L (A) [ L (B) = L (A [B).5. Is it true that if A and B are subsets of R then L (A)\L (B) = L (A \B)?Give an answer in the following cases:

(a) A and B are closed.

(b) A and B are open.

(c) A and B are intervals.

(d) General case.

6. Given that S R, S 6= ? and S bounded above but maxS does not exist,prove that supS must be a limit point of S. State and prove a similarresult for inf S.

7. Let S R. Prove that L (S) must be closed.8. Prove that if U is open then L (U) = U .