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Suppose $C$ is a subset of $\mathbb{R}$ and for any sequence of points $(x_n)$ in $C$ so that $(x_n)$ converges. Suppose $\lim(x_n) = x$ which is an element of $C$. Is $C$ closed? Why or why not?

I would think it has to be closed if $(x_n)$ converges correct?

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3 Answers 3

If all convergent sequences of points of $C$ have their limits in $C$, then yes, $C$ must be closed. You can prove it by showing that $\Bbb R\setminus C$ is open. If $\Bbb R\setminus C$ is not open, there is a point $x\in\Bbb R\setminus C$ such that for every $\epsilon>0$, $(x-\epsilon,x+\epsilon)\nsubseteq\Bbb R\setminus C$, or in other words, $(x-\epsilon,x+\epsilon)\cap C\ne\varnothing$. Thus, for each $n\in\Bbb Z^+$ there is a point

$$x_n\in\left(x-\frac1n,x+\frac1n\right)\cap C\;.$$

Can you show that $\langle x_n:n\in\Bbb Z^+\rangle$ is a convergent sequence of points of $C$ whose limit is $x$? Your hypothesis would then imply that $x\in C$, contradicting our choice of $x$ and proving that $\Bbb R\setminus C$ must be open, so that $C$ must be closed.

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Where did the 1/n come from? –  Samuel Gregory Sep 19 '12 at 4:25
    
@Samuel: Out of my head. I could just as well have used $\frac1{2^n}$, or $\frac1{n^2}$. I’ll tell you exactly what I was doing if you insist, but see if you can figure out first why any of these would have worked; it has to do with proving that the $x_n$’s converge to $x$. –  Brian M. Scott Sep 19 '12 at 4:28
    
But 1/n is not convergent though is it? –  Samuel Gregory Sep 19 '12 at 4:30
    
$\frac1n\to0$ of course –  Hagen von Eitzen Sep 19 '12 at 4:33
    
oh wow, I was thinking of something else. I see what you mean –  Samuel Gregory Sep 19 '12 at 4:34

Let $x$ be in the closure of $C$. There exists $x_n \in C$ such that $|x_n - x| < 1/n$ for every integer $n > 0$ If $m > n$, then $|x_m - x| < 1/m < 1/n$. Hence $\lim_n x_n = x$. By the assumption, $x \in C$. Hence $C$ is closed.

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Where did 1/m and 1/n come from? –  Samuel Gregory Sep 19 '12 at 4:26
    
$n$ is every integer, i.e. $n = 1, 2, \dots$. and $m$ is any integer such that $m > n$. –  Makoto Kato Sep 19 '12 at 4:28
    
Isn't say x be in the closure of C the same as x is an element of C? –  Samuel Gregory Sep 19 '12 at 4:28
    
I understand that it's every integer, just where did the 1/n come from? –  Samuel Gregory Sep 19 '12 at 4:28
    
Not the same, since we don't know $C$ is closed yet. –  Makoto Kato Sep 19 '12 at 4:29

What you seem to be asking is this: If $C$ is a subset of $\mathbb{R}$, such that for any convergent sequence of elements of $C$, the limit is also an element of $C$, then must $C$ be closed?

The answer is Yes; topological spaces that satisfy this condition are called "sequential spaces," and $\mathbb{R}$ is a sequential space. Here's why:

Suppose that $x\in\mathbb{R}$, and that every neighborhood of $x$ intersects $C$. Then for each natural number $n$, let $x_n$ be a point in the intersection of $C$ with $(x-\frac 1n, x+\frac 1n)$. Then $x_n\to x$, so $x\in C$. Therefore $C$ must be closed.

This argument works whenever you replace $\mathbb{R}$ with a "first-countable" topological space (each point admitting a countable neighborhood basis, here given by $\{(x- \frac 1n, x+\frac 1n): n\in\mathbb{N}\}$). There are, however, sequential spaces which are not first-countable.

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