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

Question

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)

Answer 1

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

Answer 2

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.

Answer 3

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, $\bar{Y} \subset K$.