# Prove that a set consisting of a sequence and its limit point is closed

Can someone please check whether the following simple proof is "mathematical"? Is it correct, complete, rigid? Can it be simplified? I'm a complete autodidact so I'm looking for someone to give me feedback to gain experience in writing proofs... This is also my first question on MSE.

The proposition:

Let $(X, d)$ be a metric space and $x_n \to x$ where each $x_n \in X$ and $x \in X$. Let $A$ be the subset of $X$ which consists of $x$ and all of the points $x_n$. Prove that $A$ is closed in $(X, d)$.

My tentative to prove this:

We first show that all infinite sequences in $A$ converge to $x$: Let $y \in X$, $y \ne x$. Then there is some open ball $B_\epsilon(x)$ with $\epsilon < d(x,y)$ containing all but finitely many elements of $A$. As $y \notin B_\epsilon(x)$ there can be no infinite sequence in $A$ converging to $y$. Consequently all infinite series in $A$ converge to a point in $A$ which therefore must be a closed set.

Edited: As rightly pointed out in the comments, I should have written in the first sentence "...sequences with infinitely many distinct terms and which converge to some point of $X$" and the last sentence should be "Consequently all infinite sequences...".

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In your last sentence, you want to say "all infinite sequences". – Martin Argerami Mar 28 '12 at 2:18
Regarding your proof, the idea is certainly ok. The proof could be simpler, more complicated, or different, depending on your definition of closed set. Here you are using that a set is closed if it contains all of its accumulation points. This is ok, but my point is that you are using a previous result (unless that's your definition of closed set) and the proof would be different if you want to do it straight from a different definition of closed ("complement of open", for example). – Martin Argerami Mar 28 '12 at 2:21
By "infinite sequence", I think you mean "a sequence containing infinitely-many distinct elements". (The constant sequence $x_n = x_0$ for all $n$ is an infinite sequence.) – Austin Mohr Mar 28 '12 at 2:22
I'm impressed with this fast and friendly feedback. Thanks a lot! @Martin Argerami: Yes, you're right. I wanted to say "sequence" rather than "series". And you're right that I'm using a previous result in this case rather than doing the proof from (my) definitions. – jerico Mar 28 '12 at 2:24
@Austin Mohr: I thought that the sequence can contain duplicates without the proposition or proof being invalid... Is that wrong? – jerico Mar 28 '12 at 2:28

As I said in my comment, your proof is ok. But here is the way I would have done it. I'll write the proof in a more formal way, because in math you can only talk loose after you master writing properly.

Let $y\in X\setminus A$. Let $\varepsilon=d(y,x)/2$. Then, by the convergence $x_n\to x$, there exists $n_0$ such that $x_n\in B_\varepsilon(x)$ for all $n\geq n_0$. So, for $n\geq n_0$, $$\tag{1} d(x_n,y)> d(x,y)-d(x_n,x)>d(x,y)-\varepsilon=d(x,y)/2.$$ Let $\delta=\min\{d(x,y),d(x_1,y),\ldots,d(x_{n_0},y)\}/2$. Then $d(y,x_n)\geq\delta$ if $n\leq n_0$, and by ($1$) $d(y,x_n)\geq\delta$ if $n\geq n_0$. This shows that $B_\delta(y)$ has no intersection with $A$, i.e. $B_\delta(y)$ is contained in $X\setminus A$. As $y$ was arbitrary, this shows that $X\setminus A$ is open, i.e. $A$ is closed.

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Great answer. That's exactly what I was looking for. :-) – jerico Mar 28 '12 at 19:11

What you really want to do is verify that $A$ satisfies your definition of closed set: that first of all any sequence $y_n \in A$ has a limit, i.e. that there exists $y$ so that $\lim y_n = y$, and second of all that $y \in A$. Informally speaking you want to say that every sequence in $A$ converges to a limit that is also in $A$ (note that the limit does not need to be $x$). This essentially eliminates the problems people pointed out in the comments. Based on this I bet you can write a new proof pretty quickly. Nevertheless, here is a quick outline of a solution:

If $y_n \in A$ you know that $y_n = x_{n_k}$ or $y_n = x$. From this you may conclude that either $y_n$ is eventually constant or $y_n$ converges to $x$. Either way you see that $y_n$ converges to a limit contained in $A$.

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Thanks @treble for this additional approach. If I could choose two answers then I would have chosen yours, too. I slightly preferred the other answer because in my special situation it was useful to have the proof spelled out in detail. – jerico Mar 28 '12 at 19:01
It is false that every sequence in A converges! Consider the sequence x_1, x_2, x_1, x_2, x_1, x_2, x_1, x_2, ... – Omar Antolín-Camarena Mar 28 '12 at 23:21
Of course, it doesn't matter that not every sequence in A converges, because the right definition of closed is not "every sequence in A converges to a point of A", but rather "every sequence in A which happens to converge to a point of X actually converges to a point of A". – Omar Antolín-Camarena Mar 28 '12 at 23:22
Yes, of course you are right- no need for the exclamation point. I should have said that every sequence has a subsequence that converges, which is what I was thinking, or I should have said that every convergent sequence converges to a point in $A$. – treble Mar 29 '12 at 1:15