Compactness of $\{X_n\}\cup\{v\}$ in a normed vector space

Question

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

I'm studying for a final and this problem continues to give me problems. While I have the answer I'm looking for as much detail as possible so I can understand the concept/how to prove this without using open covers.

Thanks!

Answer 1

Compactness means that every open cover has a finite subcover. And, the key idea is this: convergence of $(X_n)$ to $v$ tells us that for any definition of 'close', all but finitely many elements of $(X_n)$ must lie 'close' to $v$. We can use this to make a set in an open cover that covers $v$, also cover all but finitely many other points; then, we can be heavy-handed in choosing cover sets that take care of these last few stragglers.

To be formal:

Let $\mathcal{A}$ be an open cover of $S$. Since it covers $S$, there must be some $A_v\in\mathcal{A}$ such that $v\in B$.

Since $A_v$ is open and $v\in A_v$, there must exist some $\epsilon>0$ so that $B_v(\epsilon)\subseteq A_v$, where $B_v(\epsilon)$ is the ball of radius $\epsilon$ about $v$.

But, because $(X_n)$ converges to $v$, there must exist $N\in\mathbb{N}$ so that $\lvert X_n-v\rvert<\epsilon$ for all $n>N$; hence for $n>N$, $X_n\in B_{\epsilon}(v)\subseteq A_v$.

So, the only elements of the sequence $(X_n)$ that may not be covered by $A_v$ are $X_1,\ldots, X_N$. For each of these $X_i$, there must be some $A_i\in\mathcal{A}$ such that $X_i\in A_i$.

Then $\{A_v, A_1,\ldots,A_N\}$ is a finite subcover of $\mathcal{A}$.

Since this can be repeated for any open cover, the space must be compact.

Answer 2

Suppose you have an open cover $\bigcup_\alpha U_\alpha$ of $S$.

$v$ is covered by some open set in the open cover, call it $U_1$.

Since $U_1$ is open, there is some $\epsilon>0$ such that $B_\epsilon(v)$ (the open ball of radius $\epsilon$ centered at $v$) lies entirely in $U_1$, i.e. $B_\epsilon(v) \subset U_1$.

By the definition of convergence, there exists some $N$ such that $X_n \in B_\epsilon(v) \subset U_1$ for all $n \ge N$.

So $U_1$ covers all points of $S$ except possibly $X_1,\ldots,X_N$. You can add $\le N$ more sets from the open cover to cover the rest.

Answer 3

The intuition here is that the only place a sequence in $S$ can accumulate is near $s$, so if we have a sequence in $S$, either some of it accumulates near $s$ or it must repeat itself infinitely often. Let's formalize.

Let us use the fact that a metric space is compact if and only if every sequence in it has a convergent subsequence. Let $y_n$, $n=1,2,\ldots$ be a sequence in the set $S$.

Notice that each $y_n = x_{m_n}$. We have two possibilities: either there is some $N$ such that $m_n < N$ for all $n$ or there isn't.

In the first case, then for every $N$ we have some $n$ such that $m_n > N$. This means that we can find a subsequence "walking out to the tail." Such a subsequence must accumulate to $s$ - use the definition of the convergence of the $x_n$ to show this.

In the second case, then there exists such an $N$. Convince yourself that there must exist an index $m<N$ such that $y_n = x_m$ for infinitely many $n$. We have just produced a convergent subsequence of $y$s, do you see why?

Spoilers below.

Let $y = \{y_n\}$ be a sequence in $S$. We show that $y$ has a convergent subsequence. We first dispense with the case that $s$ is contained in $y$. Either $y_n = s$ for infinitely many $n$, in which case the subsequence consisting of such $y_n$ is a convergent subsequence, or not. In the latter case, pass to the subsequence of all $y_n\neq s$. Now for each $n$, we have $y_n = x_{m_n}$ is some element of the sequence $\{x_k\}$. Either there exists some $N$ such that $m_n < N$ for all $n$ or there doesn't. In the former case, there must be some $m<N$ such that $y_n = x_m$ for infinitely many $n$. In this case, the subsequence of such $y_n$ is a convergent subsequence. In the latter case, then for every $N$ there exists some $m_n > N$. Define a subsequence $y_{n_k}$ by letting $n_1=1$ and then inductively choosing $y_{n_k}$ to be some $x_{m_{n_k}}$ where $m_{n_k} > m_{n_{k-1}}$, where the existence of such an $m_{n_k}$ is guaranteed by letting $N = m_{n_{k-1}}$. Now for each $\epsilon > 0$ there exists an $M$ such that for all $m>M$ we have $\|x_m - s\| < \epsilon$. By the construction of our sequence, for each $\epsilon > 0$ there exists an $M$ such that for all $n_k > M$ we have $\|y_{n_k} - s\} < \epsilon$, so we have that $y_{n_k}$ converges to $s$.