# Showing that if $(x_n) \rightarrow x$, then $\{x_1,x_2,...\}\cup{x}$ is compact

I want to show that if $(x_n)$ is a convergent sequence in a metric space, $(x_n) \rightarrow x$, then the subset $X = \{x_1,x_2,...\}\cup{x}$ is compact. Does this proof suffice?

Since $(x_n)$ converges, all of its subsequences must also converge. Now to show $X$ is compact, we need to show any sequence in $X$ has a convergent subsequence. So consider $(y_n) \subset X$. If $y_n = x$ for all $n$, we are done. Similarly, if $y_n$ eventually has infinitely many terms equal to $x$, take that subsequence.

So we are left with the case where only finitely many terms from $y_n$ are $x$. Writing $(y_n) = \{y_1,y_2,...\}$, take the subsequence given by deleting terms equal to $x$, so $y_{n_k}=\{y_1,y_2,...\}\backslash \{x\}$ This is now clearly a subsequence of $(x_n)$, so must converge. Thus $X$ is compact.

That seems too straightforward, am I missing something? Thanks!

• The proof doesn't handle the possibility of $(y_n)$ being a reordering (but not a subsequence) of $(x_n)$. While this technicality can be resolved, I think it would be a lot easier to use the open cover definition of compactness. Apr 6, 2017 at 18:24
• We haven't covered open covers in my Analysis course yet unfortunately. For resolving that technicality, would constructing a decreasing/increasing subsequence from the re-ordering work? (Not sure how to actually write that out though). Apr 7, 2017 at 9:26

Let $(y_k)$ be an arbitrary sequence of terms from $\{x_i\}_{i=1}^{\infty}\cup \{x\}$, and let $d$ be the metric for the metric space. Since $x$ is a limit point, for each $n=1,2,3,\ldots$, there are an infinite number of values of $k$ such that $d(x,y_k)<1/n$. Let $k_1$ be the smallest $k$ such that $d(x,y_{k_1})<1$. For each $i$, define $k_i$ inductively to be the smallest integer such that $k_i>k_{i-1}$ and $d(x,y_{k_i})<1/i$. The subsequence $(y_{k_i})$ converges to $x$ and is a convergent subsequence of $(y_k)$. Thus, every sequence has a convergent subsequence, as claimed.
We'll need the following property: if a sequence $$x_{n} \to x$$, then any open set that contains $$x$$ (the limit) must contain all but finitely many terms of the sequence.
So now say we have an open cover $$\{O_{i}\}$$ of $$\{x_{1}, x_{2}, \dots\} \cup \{x\}$$, find the open set $$O$$ in the open cover that covers $$x$$. By the property stated above, this open set contains all but finitely many points of the sequence. Let $$x_{1}, \dots, x_{N}$$ be the points that it (potentially) does NOT contain. Now we consider the open sets $$O_{1}, \dots, O_{N}$$ where $$x_{i} \in O_{i}$$ and then the collection $$O \cup \bigcup_{i=1}^{N} O_{i}$$ is a finite subcover of the open cover that contains all points in $$\{x_{1},x_{2}, \dots\} \cup \{x\}$$. So, in this way, we can find a finite subcover for any open cover of the set, i.e. the set is comapct!