Prove the closure is closed and is contained in every closed set

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)

• What is $A$? Do you mean $Y$ ? Apr 25, 2018 at 17:44
• @PrasunBiswas Thanks, this was a typo. I corrected it.
– mdcq
Apr 25, 2018 at 17:54
• 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., $A=A^{\circ}$ ? Apr 25, 2018 at 18:00
• @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.
– mdcq
Apr 25, 2018 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 $$B_{\varepsilon'}(y)\subset X\setminus Y$$, 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 $$F\supset Y\cup\partial Y=\overline Y$$.
• Clear as always. Thank you!
– mdcq
Apr 25, 2018 at 18:43
• You must correct the fifth line. It's X\Y not Y\X. Nov 11, 2022 at 9:41
• @CHOSM Done. Thank you. Nov 11, 2022 at 10:06

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.