Compact susbset in normed vector space

[burritoloco]

b]1. Homework Statement [/b]
Let E be a normed vector space. Let (x_n) be a convergent sequence on E and x its limit. Prove that A = {x}U{x_n : n natural number} is compact.

Homework Equations

A is compact iff for any sequence of A, it has a cluster point, say a in A, i.e. there is a subsequence converging to some a in A and
for any ball around a there are an infinite # of elements from the arbitrarily chosen sequence in A.

The Attempt at a Solution

I wrote a proof, but for some reason I'm not too confident about it. Could anyone check it please?

Let (y_m) be a sequence of A, and let e>0.
Claim: The set S = {N natural number : there exists m >= N s.t. y_m in B(x,e)} is infinite.
Proof:
If (y_m) = (x_n) we are done. Suppose on the contrary that it is finite. Let M = max(S), where M = 0 if S is empty. We get by the definition of S:
for all N > M for all m >= N y_m not in B(x,e).

We get an infinite # of y_m s.t. y_m not in B(x,e). Whenever y_m != x, then y_m = x_n, some n; thus there exist m,n >= N s.t. y_m = x_n (otherwise, if there are infinitely distinct y_m's, then there are infinite natural numbers n s.t. n < N - impossible. Else if there are finitely distinct y_m's, then (y_m) is convergent in A and thus it has a cluster in A) and x_n not in B(x,e). We get
for all N > max(S) there exists n >= N x_n not in B(x,e).

This contradicts that x is the cluster of (x_n) iff there exists N s.t. for all n >= N x_n in B(x,e). q.e.d.

So now we get an infinite subsequence y_mk with each y_mk in B(x,e). This implies that x is a cluster of (y_m). Moreover x in A. Thus A is compact. q.e.d.

 

[mighty2000]

Your proof is correct! You have used the definition of compactness and the fact that a convergent sequence has a cluster point to show that A is compact. Well done!