Prove: $F=\{x_n\}\cup \{x\}$ Is closed

Question

Let $M$ be a metric space, $x_n\in M$ a sequence which converges to $x\in M$

Prove: $F=\{x_n\}\cup \{x\}$ is a closed set

So we have $x_n\to x$ such that $x_n\in F$ and $x\in F$ and we know that a set is closed if it contains all of its accumulation points, so $F$ is closed

Or must I look at $F^c$ and prove that it is open?

Answer 1

You can also show that $F^c$ is open, let $y\in F^c$, suppose that for every $n>0$, there exists $x_{n_p}$in the open ball $B(y,{1\over n})$, this implies that $lim_nx_{n_p}=y=x$ contradiction. Thus there exists $n_0$ such that $B(y,{1\over n_0})$ does not contain any $x_n$, it does not contain $x$ also, we deduce that $F^c$ is open.

Answer 2

How you prove a set is "closed" depends strongly on what definition of "closed" you are using. One commonly used definition is that a set is closed if and only if it contains all of its limits points. Show that the only limit point of this set is x.

Answer 3

Here's a direct proof without considering the compliment of the set. This proof also works for an arbitrary Hausdorff space -- not just a metric space. Suppose $\{ y_n \}$ is a sequence in $F$ and $y_n \rightarrow y \in M$. If $ \{ y_n \} $ is eventually constant, we're done. So suppose that the sequence is not eventually constant.

Claim: $\{y_n\}$ has a subsequence $\{y_{n_j} \}$ such that $m_1 < m_2 < \cdots$, where $y_{n_j} = x_{m_j}$

To see this, suppose there were no such subsequence. Then there is some $M$ such $\{y_n\} \subset \{x_1, x_2, \cdots, x_M \}$. But this means that $\{y_n\}$ is convergent if an only if it is eventually constant. We already have said this is not the case.

Using the above claim, we have a subsequence $\{y_{n_j}\}$ and by the condition on the indices, $\{y_{n_j}\}$ is also a subsequence of $\{ x_n \}$. Since a subsequence of a convergent sequence converges to the same limit, we have that $y_{n_j} \rightarrow x$ and $y_{n_j} \rightarrow y$. Since limits are unique in a metric space, we have $y=x$.

Therefore, for both cases we have that $y_n \rightarrow y \in F$, making $F$ a closed set.