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I have a question that I am trying to answer:

Suppose $C$ is a subset of the real numbers and $C$ is closed. If $\{x_n\}$ is a sequence of points in $C$ and $\lim x_n =x$, is $x$ an element of $C$? Why or why not?

I am thinking I should use either def. of open: $O$ is a subset of $\Bbb R$ is open if $x$ is an element of $O$, then there exists $\epsilon_x > 0$ such that $(x-\epsilon_x, x+\epsilon_x)$ is a subset of $O$

$C$ is a subset of $\Bbb R$ is closed if $C^C$ is open

3 Answers 3

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The definition of a closed set is a set that contains all of its limit points. Why don't you just use this definition?

To your credit, the two books I've seen for this sort of thing do use different definitions. They include the other results as theorems.

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You can do it that way. Assume, in order to get a contradiction, that $x\notin C$. Then there is an $\epsilon>0$ such that $(x-\epsilon,x+\epsilon)\subseteq\Bbb R\setminus C$. What does this tell you about $(x-\epsilon,x+\epsilon)\cap C=\varnothing$? Can you get a contradiction from here?

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    I understand the first half. That is just stating the definition of open. I am trying to understand what contradiction is yielded by looking at the intersection. Shouldn't we be looking at the limit?2012-09-19
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    @Samuel: What does it mean to say that $x$ is the limit of the sequence $\langle x_n:n\in\Bbb N\rangle$? For every $\epsilon>0$ there is ... ?2012-09-19
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    there is a K(epsilon) in N such that n>K(epsiolon), then abs(xn-x)2012-09-19
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    @Samuel: Right, so for each $n>K(\epsilon)$ you’ll have $x_n\in(x-\epsilon,x+\epsilon)$. But $x_n\in C$, so ... ?2012-09-19
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    I am sorry, I am confused. How do you know xn is an element of C? In the beginning, you said x is not an element of C.2012-09-19
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    Your hypothesis is that the sequence lies in the set $C$ and that $x$ is its limit. We want to prove that $x\in C$, so we assume that it isn’t and try to get a contradiction. In getting that contradiction, we’ll use the hypothesis that the points $x_n$ **are** all in $C$.2012-09-19
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    Is the other proof correcT? It seems to me that the def. of open was used wrongly.2012-09-19
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Let $U$ be the complement of $C$. Suppose $x \in U$. Since $U$ is open, there exists $\epsilon > 0$ such that $y \in U$ whenever $|y - x| < \epsilon$. Since $\lim x_n = x$, there exists an integer $n$ such that $|x_n - x| < \epsilon$. Hence $x_n \in U$. This is a contradiction. Hence $x \in C$.

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    I'm sorry, where is the contradiction? You said that x was an element of U at the beginning and then got it back in the end right?2012-09-19
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    The contradiction is $x_n \in U$.2012-09-19
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    Wait a second, did the abs(y-x) < epsilon come from def of limit? It appears you say it came from it being open.2012-09-19
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    @Samuel: The statement that there is an $\epsilon>0$ such that $y\in U$ whenever $|y-x|<\epsilon$ comes from the fact that $U$ is open and the hypothesis that $x\in U$. This is the same $\epsilon$ as in my version. Makoto’s $n$ is then any integer greater than your $K(\epsilon)$: for every $n>K(\epsilon)$, $|x_n-x|<\epsilon$. But then by the choice of $\epsilon$ we know that $x_n\in U$, contradicting the assumption that $x_n\in C$. It’s really exactly the same as the argument that I was prompting you to construct.2012-09-19
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    I think I just need to read up more on this topic. I am not following where abs(y-x) < epsilon comes from. I know that abs(xn-L)2012-09-19
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    @SamuelGregory How do you define an open subset of $\mathbb{R}$?2012-09-19
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    I defined it in my original problem: O is a subset of R is open if x is an element of O, then there exists epsilon_x > 0 such that (x-epsilon_x, x+epsilon_x) is a subset of O2012-09-19
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    @SamuelGregory That's a correct definition. Since $U$ is open and $x \in U$, there exists $\epsilon > 0$ such that $(x - \epsilon, x +\epsilon) \subset U$, which means that $y \in U$ whenever $|y - x| < \epsilon$.2012-09-19
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    Could you just quickly explain how the first part implies the second part?2012-09-19
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    @SamuelGregory $y \in (x - \epsilon, x +\epsilon)$ if and only if $|y - x| < \epsilon$.2012-09-19