Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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

share|cite|improve this question
up vote 1 down vote accepted

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 $x \in C$.

share|cite|improve this answer
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? – Samuel Gregory Sep 19 '12 at 3:16
The contradiction is $x_n \in U$. – Makoto Kato Sep 19 '12 at 3:20
Wait a second, did the abs(y-x) < epsilon come from def of limit? It appears you say it came from it being open. – Samuel Gregory Sep 19 '12 at 3:25
@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. – Brian M. Scott Sep 19 '12 at 3:33
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)<epsilon comes from the def of limit, not from being open – Samuel Gregory Sep 19 '12 at 3:37

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?

share|cite|improve this answer
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? – Samuel Gregory Sep 19 '12 at 3:09
@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 ... ? – Brian M. Scott Sep 19 '12 at 3:12
there is a K(epsilon) in N such that n>K(epsiolon), then abs(xn-x)<epsilon – Samuel Gregory Sep 19 '12 at 3:14
@Samuel: Right, so for each $n>K(\epsilon)$ you’ll have $x_n\in(x-\epsilon,x+\epsilon)$. But $x_n\in C$, so ... ? – Brian M. Scott Sep 19 '12 at 3:17
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. – Samuel Gregory Sep 19 '12 at 3:19

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.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.