Let $(X,d)$ be a metric space and $Y \subseteq X$ a subset. I want to show the following without using accumulation points or limit points at all.

  1. $\overline Y$ is a closed subset of $X$.
  2. $\overline Y$ is contained in every closed set which contains $Y$.

My definitions are:

The interior of $Y$ is

$$Y^\circ = \{y \in Y \mid \exists \varepsilon \gt 0:B_\varepsilon(y) \subseteq Y\}.$$

The boundary of $Y$ is

$$ \partial Y= \{x \in X \mid \forall \varepsilon \gt 0: B_\varepsilon(x) \cap Y \neq \emptyset \neq B_\varepsilon(x) \cap (X \setminus Y) \}.$$

And the closure of $Y$ is

$$\overline Y = Y \cup \partial Y.$$

I know also that a set $O \subseteq X$ is open in $X$ iff every convergent sequence with limit in $O$ has also almost all of its terms in $O$. And a set $A$ is closed in $X$ iff every convergent sequence with all of its terms in $A$ has also its limit in $A$.

I tried showing that $X \setminus \overline Y$ is open as well as using the above result for sequences but both times I got stuck in complicated set expressions resulting from figuring out what $X \setminus \overline Y$ is. Can you show me how to prove this?

I found only proofs using limit or accumulation points (links: here, here, here, here, here)

  • $\begingroup$ What is $A$? Do you mean $Y$ ? $\endgroup$ – Prasun Biswas Apr 25 '18 at 17:44
  • $\begingroup$ @PrasunBiswas Thanks, this was a typo. I corrected it. $\endgroup$ – philmcole Apr 25 '18 at 17:54
  • $\begingroup$ What is your definition of "closedness"? Is it that the complement is open with the definition of "open" being that all its points are interior points, i.e., $A=A^{\circ}$ ? $\endgroup$ – Prasun Biswas Apr 25 '18 at 18:00
  • $\begingroup$ @PrasunBiswas No. We defined open set as containing an $\varepsilon$-ball for every of its points and closed sets are defined as having their complement open. $\endgroup$ – philmcole Apr 25 '18 at 18:20
  1. Asserting that $\overline Y$ is closed is the same thing as asserting that $X\setminus\overline Y$ is open. Take $x\in X\setminus\overline Y$. Since $x\notin\partial Y$, there is some $\varepsilon>0$ such that $B_\varepsilon(x)\cap Y=\emptyset$ or that $B_\varepsilon(x)\cap(X\setminus Y)=\emptyset$. But the second possibility cannot occur, since $x\in B_\varepsilon(x)\cap(X\setminus Y)$. Therefore, $B_\varepsilon(x)\cap Y=\emptyset$. Now, take $y\in B_\varepsilon(x)$ and take $\varepsilon'>0$ such that $B_{\varepsilon'}(y)\subset B_\varepsilon(x)$. Then $B_{\varepsilon'}(y)\subset Y\setminus X$, which proves that $y\notin\partial Y$. Therefore, $B_\varepsilon(x)\cap(Y\cup\partial Y)=\emptyset$. In other words, $B_\varepsilon(x)\cap\overline Y=\emptyset$. This proves that $B_\varepsilon(x)\subset X\setminus\overline Y$ and, since this happens for each $x\in X\setminus\overline Y$, $X\setminus\overline Y$ is open.
  2. Let $F$ be a closed set such that $F\supset Y$. Let $y\in\partial Y$. Can we have $y\notin F$? No, because $X\setminus F$ is open and therefore there would be a $\varepsilon>0$ such that $B_\varepsilon(y)\subset X\setminus F$. In other words, $B_\varepsilon(y)\cap F=\emptyset$ and, in particular, $B_\varepsilon(y)\cap Y=\emptyset$. This is absurd, since $y\in\partial Y$. So, this proves that $F\supset Y\cup\partial Y=\overline Y$.
  • $\begingroup$ Clear as always. Thank you! $\endgroup$ – philmcole Apr 25 '18 at 18:43

Let $a\in\bar{Y}^c = Y^c \cap \partial Y^c$. Claim is $\text{dist}(a, Y)>0$. Otherwise, show that it is either in $Y$ or in else in $\partial Y$. When you have this, choose $\epsilon$ to satisfy being in the interior.

Let $Y \subset K$, where $K$ is closed. Then $Y^c \supset K^c$. Note that ${Y^{c}}^{\circ} \supset K^c$ since $K^c$ is open. Therefore, $\bar{Y} \subset K$.


Let $\{x_n\}_n$ be a convergent sequence in $X$ with limit $x\in X\setminus\overline Y$. In particular $x\notin\partial Y$, hence there exists $\epsilon>0$ such that one of $B_\epsilon(x)\cap Y$, $B_\epsilon(x)\cap (X\setminus Y)$ is empty. Of course the latter is wrong for $x\notin Y$. Hence $B_\epsilon(x)\cap Y=\emptyset$. Note that almost all $x_n$ are in $B_\epsilon(x)$ (and consequently in $X\setminus Y$). And for each such $x_n$, we have $B_{\epsilon-d(x_n,x)}(x_n)\subseteq B_\epsilon(x)$ and hence $B_{\epsilon-d(x_n,x)}(x_n)\cap Y=\emptyset$. We conclude that $x_n\notin \partial Y$. Together with $x_n\notin Y$, we get $x_n\in X\setminus \overline Y$, as desired.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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