# Real Analysis - Closed and Open Sets

I have the next question from Rudin and want to know if it was answered correctly: Is every point of every closed set E $\subset$ $R^2$ a limit point of E?

Consider the interval $E=[0,1]$ which is closed. By definition, every point of $E$ is a limit point. It is proved that all points in $R^2$ are limit points. Therefore $E^c$ must contain all limit points. But $E^c$ by definition is open ($E^c=R^2\setminus E$). This is a contradiction because $E^c$ has 2 limit points, namely $E=[0,1]$. Therefore E contains points which are not limit points.

• I don't quite get why you claim that $E^c$ must contain all limit points. It's an open set, after all. Commented May 17, 2013 at 13:30
• I confess I did not understand your argument. "It is proven that all points in $\mathbb{R}^2$ are limit points" - how did you get this conclusion? And are they limit point of $E$ or limit points of $\mathbb{R}^2$? Commented May 17, 2013 at 13:30
• WHat is the definitions for closed sets you have? Commented May 17, 2013 at 13:31
• You mention Rudin and the definition given is: E is closed if every limit point of E is a point of E. This is different to saying that E is closed if every point of E is a limit point of E. For example, $E = [0,1] \cup \{2\}$ is closed using the definition given by Rudin, but using your definition it isn't closed as $2$ is not a limit point of $E$, but $2 \in E$.
– user70962
Commented May 17, 2013 at 13:41
• I think I went completely on the wrong way trying to prove it with the interval [0,1]. To why all points in $R^2$ are limit points, I also read that and used it as an argument. Sorry I didn't mention that. Commented May 17, 2013 at 13:54

No, your argument is not correct. I’ll comment on it in detail.

Consider the interval $E=[0,1]$ which is closed. By definition, every point of $E$ is a limit point.

While it is true that every point of $[0,1]$ is a limit point of $[0,1]$, this is not a matter of definition: there is nothing in the definition of closed set that requires every point of a closed set to be a limit point of that set. Moreover, $[0,1]$ isn’t a subset of $\Bbb R^2$. I suspect that you’re thinking of the set $[0,1]\times\{0\}$, the set of points on the $x$-axis between $0$ and $1$ inclusive.

It is proved that all points in $R^2$ are limit points.

Limit points of what? They’re all limit points of $\Bbb R^2$, but that has nothing to do with $E$, which isn’t even a subset of $\Bbb R^2$. And if your $E$ was really supposed to be $[0,1]\times\{0\}$, then it’s not true that all points of $\Bbb R^2$ are limit points of $E$: if they were, they’d have to be in $E$, since $E$ is closed, and a closed set contains all of its limit points.

Therefore $E^c$ must contain all limit points.

Again, all limit points of what? And what does $\Bbb R^2$ have to do with $E^c$? $E^c=(\leftarrow,0)\cup(1,\to)$, a subset of $\Bbb R$, not of $\Bbb R^2$. For that matter, why are you bringing in $E^c$ at all? The question is whether every point of $E$ is a limit point of $E$, and $E^c$ isn’t going to help you to answer that question.

But $E^c$ by definition is open ($E^c=R^2\setminus E$).

See above: $E^c=\Bbb R\setminus E$, not $\Bbb R^2\setminus E$.

This is a contradiction because $E^c$ has 2 limit points,

No, every point of $(\leftarrow,0]\cup[1,\to)$ is a limit point of $E^c$; that’s far more than two limit points!

namely $E=[0,1]$.

No, the only points of $E$ that are limit points of $E^c$ are $0$ and $1$; the points of $E$ that are in $(0,1)$ are not limit points of $E^c$.

Therefore E contains points which are not limit points.

This doesn’t follow from anything that you’ve said, and in fact it contradicts your very first claim about $E$ up above. Moreover, it isn’t true, assuming that you’re talking about limit points of $E$: every point of $E$ is a limit point of $E$.

In order to answer the question, you must do one of two things: either prove that whenever $E$ is a closed subset of $\Bbb R^2$, then every point of $E$ is a limit point of $E$, or find an example of a closed subset $E$ of $\Bbb R^2$ containing a point that is not a limit point of that set $E$. You won’t be able to do the former, because it isn’t true. For the latter, just take any non-empty finite subset of $\Bbb R^2$, for instance $E=\{\langle 0,0\rangle\}$: $E$ is closed, and it has no limit points at all. In particular, $\langle 0,0\rangle$ is not a limit point of $E$, because it is not true that every open neighborhood of $\langle 0,0\rangle$ contains a point of $E$ different from $\langle 0,0\rangle$. If you want $E$ to be an infinite set, you can use $E=\big(\Bbb R\times\{0\}\big)\cup\{\langle 0,1\rangle\}$, for instance. This is a closed set, but $\langle 0,1\rangle$ is a point in $E$ that is not a limit point of $E$: if $\epsilon\le 1$, the open ball of radius $\epsilon$ centred at $\langle 0,1\rangle$ does not contain and point of $E$ besides the point $\langle 0,1\rangle$ itself. For a more extreme example, let $E=\{\langle m,n\rangle:m,n\in\Bbb Z\}$. You should try to prove that $E$ is an infinite closed set that has no limit points at all.

A closed set always contains all of its limit points. That is, suppose that $E$ is closed. If $p$ is a limit point of $E$, then $p\in E$. The point of this exercise is to show that the converse is not true: there can be a point $p\in E$ that is not a limit point of $E$.

The answer is no. Take a subspace of $\mathbb{R}^2$ consisting of a single point $P$. Then $\{P\}$ is closed, since its complement $\mathbb{R}^2\setminus\{P\}$ is open in the usual topology of $\mathbb{R}^2$. Now, $P$ is a point of $\{P\}$ but not a limit point for $\{P\}$ since you can't define in any way $P$ as the limit of a sequence of points of $\{P\}$ different from the constant sequence.

• I'm upvoting you because this doesn't deserve a negative score, but it doesn't address OP's misunderstandings about limit points. Commented May 17, 2013 at 14:10