Every closed interval in $R^1$ is closed set (check logic) I want to prove, that every closed interval in $\mathbb{R}^1$ is closed set.
Can I use the argument, that there are uncountably many points in $\mathbb{R}^1$ and if we will take arbitrary point $x$,that belongs to a closed interval,and take arbitrary neighbourhood around this  point, we will always find a point, that belongs to the interval. It means, that our point is a limit point. Because point is arbitrary and neighbourhood as well it means, that this set is closed.
 A: As far as I understand from your claim, you prove that every member of a closed interval is a limit point of the interval. You need to prove that every limit point of a closed interval is a point of the interval, instead.      
Pick any closed interval $[a,b]$. Pick any $x$ that is not a member of $[a,b]$, hence $x < a$ or $b < x$. Assume that $x$ is a limit point of $[a,b]$, hence it is not closed. Without loss of generality, pick $b < x$. Then in the segment $(b,x)$ there are no points of $[a,b]$, contradicting $x$ being a limit point. Then there is no point which is not a member of the closed interval and is a limit point of the interval in the same time.
With your proof and this, additionally it becomes apparent that $[a,b]$ is a perfect set.
A: You're supposed to prove that any limit point of $I$ is within $I$, but what you do in your proof is to a-priori assume that $x\in I$ and then prove that $x$ is a limit point of $I$. Actually what you're proving is that $I$ is dense in itself, note how your proof also works for open intervals.
Besides your proof is broken as the uncountability of a $\mathbb R$ doesn't imply that at all. First it doesn't say anything about $I$ being uncountable, and even if it were one can construct such sets that are not dense in itself. Note that a closed interval as opposed to an open does not need to have infinite elements: the closed interval $[0,0]$ only contains $0$.
What you instead should have done is to start with a limit point $x$ of $I=[a,b]$. Which in turn mean that every neighborhood $(x-\epsilon, x+\epsilon)$ of $x$ intersects $I$, that there is a point $c\in I$ such that $|x-c|<\epsilon$. Now $a\le c\le b$ which means that $x<b+\epsilon$ and $x>a-\epsilon$. This has to be true for any $\epsilon>0$ which means that $a\le x\le b$ (for assume for example that $x<a$ then we would have $a-x>0$ and the assumption $x>a-\epsilon$ would fail if $\epsilon = (a-x)/2$ since then $a-\epsilon = a-a/2+x/2 = a/2+x/2 < x$).
A: I don't see how your argument is in any way a proof of the original claim. There are two major problems with it:


*

*You did not tell us what definition of closed set you are using.

*You did not structure the argument, so it is impossible to follow.



A good argument would look something like this:

Claim: Every closed interval $I$ is a closed set.
Proof: Let $I$ be a closed interval.
  By definition, a set $A$ is closed if and only if ___________.  Therefore, we want to prove that ______________ is true for $I$. ________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________
  Therefore, we have shown that ________________ is true for $I$. We conclude that $I$ is closed.
Because $I$ was an arbitrary closed interval, we know that every closed interval is a closed set.

A: As the other answers have done a good job of explaining why your reasoning needs some work, I'll only discuss your original problem here.
Let $A$ be a subset of $\mathbb{R}$. There are two common definitions of a closed set you should be familiar with:


*

*The set $A$ contains all of its limit points. Symbolically, this means that $A=\bar{A}$.


*The complement, $\mathbb{R}\setminus A$, is open.

We will prove that $A=[a,b]$ for any $a,b\in\mathbb{R}$ where $a<b$ is closed (in the usual topology on $\mathbb{R}$) using both definitions. I encourage you to try and spot the similarities between these two arguments.

*

*First, suppose that $x\in\mathbb{R}$ is a limit point of $A$ such that $x\notin A$. This means that for any $\epsilon>0$, the set $(x-\epsilon,x+\epsilon)$ contains points of $A$. If $x<a$, we consider any $\epsilon < \frac{a-x}{2}$. However, this implies that $(x-\epsilon,x+\epsilon)$ does not contain any points of $A$, which is a contradiction. Therefore, $x$ cannot be a limit point of $A$, so $x\in A$. By definition, this means $A$ is closed. (You should prove the case when $x>b$ yourself).
 

*Consider the complement of $A$, which we denote $B:=(-\infty, a)\cup(b,\infty)$. Given any $x\in B$, we can find some open ball $(x-\epsilon,x+\epsilon)$ that is entirely contained within $B$ (you should prove this statement). By definition, this implies that $B$ is open, and therefore $A$ is closed also by definition.

A: There is a standard definition of closed set,"the complement of an open set is called  $closed$".Any closed interval $[a,b]$ is the complement of the union of two open sets  $(-\infty,a)$ and $(b,\infty)$(union of open sets is open).
A: Equivalently, a subset A of a metric space X is closed if and only if the limit of every convergent sequence in A is in A.
Let $[a,b]$ be a closed interval. Let $(x_n)$ be a sequence in A so that $x_n$ $\rightarrow x$ where x $\notin A$.
Consider the two cases
$x<a$ and $x>b$ and try to get a contradiction in both cases by choosing a suitable $\epsilon>0$ in either case.
