I need a proof for both parts in picture: Here is the typed question that you sh

Question

I need a proof for both parts in picture:

Here is the typed question that you should see in the picture:

a) Let E be a subset of R. A point a in R is called a cluster point of E if E intersects (a-r,a+r) contains infinitely many points for every r>0. Prove that a is a cluster point of E if and only if for each r>0, E intersects (a-r,a+r){a} is nonempty.

NOTE: I SAID INTERSECT BECAUSE I COULDN'T FIGURE OUT HOW TO TYPE THE UPSIDE DOWN U. ALSO {a} MEANS WITHOUT {a}

b) Prove that every bounded infinite subset of R has at least one cluster point

Explanation / Answer

(a)

Observe that from the definition of a cluster point a is a cluster point of E => ( E intersection (a-r,a+r){a} ) is nonempty for all r>0

This is because if not, then ( (a-r,a+r) intersection E ) contains only one point a, a contradiction to the assumption that a is a cluster point.

Now conversely suppose (a-r,a+r){a} intersection E is nonempty for all r>0.

Then we need to show that (a-r,a+r) intersection E consists of infinitely many points for all r.

Suppose not,then there exists r where this is not true, then take r' = min { |a-b| : b is in E intersection (a-r,a+r){a} }. Then 0 < r' < r. (r'>0 as (a-r,a+r){a} intersection E is nonempty).

Then consider (a-r', a+r'){a}. This does not intersect E as otherwise if b is in the intersection b is in E intersection (a-r,a+r){a} and thereby r' <= |b-a| < r', a contradiction.

Now (a-r', a+r'){a} intersection E is empty, contradicting the hypothesis of the converse.

Hence our assumption is false. Thus (a-r,a+r) intersection E consists of infinitely many points for all r>0, i.e a is a cluster point.

(b)

S = bounded infinite set

Form a sequence as follows

pick any s1 in S and let x_1 = s1.

Then pick any s2 in S{s1} and let x_2 = s2.

..... pick any s(n+1) in S{s1,s2,...,sn} and let x_(n+1) = s(n+1).

By mathematical induction we get a sequence x_n of real numbers, all of which are distinct such that x_i is in S.

Now S is bounded. So x_n is in [-K,K] for some N>0.

Now [-K,K] is sequentially compact.

So x_n has a converging subsequence , call x_k_n.

Suppose this converges to s.

Then s is a cluster point of S, as given any r>0, (s-r,s+r){s} is nonempty since x_k_n converges to s, there exists N such that for all n>=N |x_k_n-x| < r.

Thus S has atleast one cluster point.

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