Prove if the complement of a subset $A$ of $\mathbb{R}^2$ is open, then $A$ is closed.

Question

Prove if the complement of a subset $A$ of $\mathbb{R}^2$ is open, then $A$ is closed. I will provide my attempt for verification!

For reference, I define a set $U$ in $\mathbb{R}^n$ to be open if for every $x$ in $U$, there is an $\epsilon$ small enough so that $B_{\epsilon}(x)\subseteq U$. A closed set is a set that contains all its limit points.

PF:

We want to show if $A^c$ is open, the $A$ is closed. Let $X=\mathbb{R}^2$ and $A\subseteq X$. Then, denote $A^c$ as the complement of A such that $A^c=\{x\in X|x\notin A\}$. Take an arbitrary point $x$ in $A^c$. Since $x$ is in $A^c$, we know $x$ is a limit point of $A^c$ but we also know that $\exists y \notin A^c$ such that $y$ is a limit point of $A^c$. Since these arbitrary point(s) $y$ are not in $A^c$, they must be in $A$. We have generated an arbitrary point $y$ that is not in $A^c$ but in $A$ that is a limit point of both sets. Thus we can conclude $A$ is closed since it contains all its limit points.

QED

It would be awesome if someone could validate this as correct or not and provide hints if I'm off a bit. Thanks in advance!

Answer 1

There are a number of problems with your proof. Taking it one bit at a time,

We want to show if $A^c$ is open, the $A$ is closed. Let $X=\mathbb{R}^2$ and $A\subseteq X$. Then, denote $A^c$ as the complement of A such that $A^c=\{x\in X|x\notin A\}$.

So far, just restating the problem, maybe a bit more clearly. Okay, that doesn't really hurt.

Take an arbitrary point $x$ in $A^c$. Since $x$ is in $A^c$, we know $x$ is a limit point of $A^c$...

It's true every element of an open set in $\mathbb{R}^2$ is a limit point of that set. But that's not a direct application of your definition of open, so is this a theorem you've already seen proved, or can you see why it must be true?

... but we also know that $\exists y \notin A^c$ such that $y$ is a limit point of $A^c$

This is not certain. The main premise of this proof should be that $A^c$ is open. How can you say anything about the limit points of an open set, since your definition of open involves open balls surrounding points, and it's the definition of closed that mentions limit points?

In particular, remember that the entire set $\mathbb{R}^2$ is an open set in $\mathbb{R}^2$. And if $A^c = \mathbb{R}^2$, it's not true that there is a limit point $y$ of $\mathbb{R}^2$ which is not an element of $\mathbb{R}^2$.

Since these arbitrary point(s) $y$ are not in $A^c$, they must be in $A$.

Okay, except be careful about switching between singular and plural. You originally claimed there exists one such point $y$. Saying there exists an object that satisfies a predicate does mean there might be just one such object or more than one, but if you need to deduce anything else from that claim, the one object needs to be enough.

The word "arbitrary" doesn't have enough context here to be meaningful. It normally means there was some earlier choice, and then you showed that in some sense that choice didn't matter. But if only one point satisfied the property of $y$, and you didn't say anything otherwise, there was no choice in the first place.

We have generated an arbitrary point $y$ that is not in $A^c$ but in $A$ that is a limit point of both sets.

You claimed $y$ is a limit point of $A^c$, but now how do you also conclude $y$ is also a limit point of $A$? The only similar thing I see was that $x$ is a limit point of $A$, but you haven't established any relationship between $x$ and $y$.

Thus we can conclude $A$ is closed since it contains all its limit points.

You claimed something about one point $y$, not something about every limit point of $A^c$. So $A$ being closed doesn't follow.

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So what is a good approach here?

The statement to prove is "if $A^c$ is open, then $A$ is closed". Let's start by expanding those with the definitions:

• $A^c$ is open means that $\forall x \in A^c : \exists \epsilon >0 : B_\epsilon(x) \subseteq A^c $.

• $A$ is closed means that every limit point $y$ of $A$ is an element of $A$.

Probably the simplest thing from this point is a proof by contradiction: Suppose $A^c$ is open and $A$ is NOT closed.

• $A$ is NOT closed means that there exists some limit point $y \in X$ of $A$ such that $y \notin A$.

But if $y \notin A$, then $y \in A^c$. Since the meaning above for "$A^c$ is open" shows a property for every element of $A^c$, this property must apply to $y$ in particular, so we can plug in $y=x$ to that property:

• $ \exists \epsilon>0 : B_\epsilon(y) \subseteq A^c $

Can you take it from here and find the contradiction?

Answer 2

I do not understand how you can postulate the existence of $y \notin A^c$ such that $y$ is a limit point of $A^c$ just from the fact that $A^c$ is open (i.e., just by using your definition of openness). I think you may be implicitly using the fact that $A$ is closed, and as such it contains all its border points, which are also the border points of $A^c$; i.e. you are using the thesis to prove the thesis. Furthermore, you it is not immediately true that if $y$ is a limit point for $A^c$ then it is also a limit point for $A$: what if $y$ is an isolated point in $A$? Any ball centered in $y$ will intersect $A^c$, but small enough balls will not contain points $\neq y$ in $A$, by definition of isolated point. So you also implicitly assumed that $A$ is perfect, i.e. all its points are limit points.

Here's a simpler way. Basically, to prove that $A$ is closed, you want to prove that a point not in $A$ cannot be a limit point of $A$. Let $x$ be such a point. Then $x \in A^c$. But $A^c$ is open, so there exists a small enough ball around $x$ that is entirely contained within $A^c$. Then this ball does not contain any points in $A$, and $x$ cannot be a limit point of $A$.

Answer 3

If $A^C$ is open, then for each $x \in A^C$ there is an $\epsilon>0$ such that $B(x; \epsilon) = \{ y \in \mathbb{R}^2 \colon |x - y| < \epsilon\} \subseteq A^C$.

If $x \in \bar{A}$, then, for each $\epsilon > 0$, $B(x; \epsilon) \cap A \neq \emptyset$.

These are your definitions, I guess. Now suppose some $x \in \bar{A} $. $x$ can't be in $A^C$, for then no $\epsilon$-neighbourhood of $x$ would be contained in $A^C$, contrary to the hypothesis that $A^C$ is open. Then, if $x \in \bar{A}$, $x \in (A^C)^C = A$. Therefore, $\bar{A} = A$ and $A$ is closed.