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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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    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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    @SamuelGregory $y \in (x - \epsilon, x +\epsilon)$ if and only if |y - x| < \epsilon.2012-09-19