A problem about compactness

[tghg]

Homework Statement

Serge Lang Undergraduate Analysis Chapter Ⅷ §1 Exe4

Let{Xn} be a sequence in a normed vector space E such that {Xn} converges to v. Let S be the set consisting of all v and Xn.
Show that S is compact.

Homework Equations

None

The Attempt at a Solution

I guess that maybe it is useful to consider it from the aspect of the definition of compactness,i.e. every sequence of elements of S has a convergent subsequence whose limit is in S. But I coudn't convince that why there must be such a convergent subsequence in every sequence, you know, some sequences are not the given ones that converge.

 

[quasar987]

Here's a plan for you.

1° Show that the only accumulation point in S is v.

2° Consider a sequence {a_n} in S. Convince yourself that there can be one of two possibilities. Either {a_n} contains a finite number distinct elements or an infinity of distinct element.

3° Show that in the first case, there is at least one element of the sequence that appears an infinity of times (this defines a converging subsequence).

4° Show that in the later case, the sequence is convergent and converges to v. (defining a "trivially converging" subsequence)

 

[tghg]

Thanks.
but I still have some problems.
I can't show that
If {a_n} contains an infinity of distinct element, the sequence is convergent and converges to v.

Is the set S bounded?

 

[StatusX]

It's probably easiest to go back to the definition in terms of open covers. If U_alpha is an open cover, v is in one of the open sets, and this open set contains all but finitely many of the x_n.

 

[quasar987]

This point raised by StatusX (namely, that If U_alpha is an open cover, v is in one of the open sets, and this open set contains all but finitely many of the x_n), is also the key to your problem:

I can't show that
If {a_n} contains an infinity of distinct element, the sequence is convergent and converges to v.

The fact that v is the only accumulation point of S means that

$$\forall\epsilon>0, \ \ |S\setminus (B_{\epsilon}(v)\setminus\{v\})|<\infty$$

In words: for any radius epsilon as small as we like, there is only a finite number of elements of S outside the open ball of radius epsilon centered on v.

It follows immediately that for any sequence {a_n} with an infinity of distinct elements and for all epsilon>0, there exist an N>0 such that for all n>N, a_n $\in B_{\epsilon}(v)\setminus\{v\}$ and that is precisely the definition of convergence.

(If it weren't so, i.e. if there was an epsilon such that there is only a finite number of elements inside the epsilon-ball, then there is an infinity of them outside. But there is only a finite number of distinct elements outside the epsilon-ball, so our sequence cannot have an infinity of distinct elements: contradiction.)