Metric Spaces: Interior of a Set

[Bashyboy]

Homework Statement

Let $(X,d)$ be some metric space, and let $A$ be some subset of the metric space. The interior of the set $A$, denoted as $int A$, is defined to be $\bigcup_{\alpha \in I} G_\alpha$, where $G_\alpha \subseteq A$ is open in $X$ for all $\alpha \in I$. The closure of $A$, denoted as $\overline{A}$, is defined to be $\bigcap_{\beta \in J} F_\beta$, where $A \subseteq F_\beta$ and $F_\beta$ is closed in $X$.

I am asked to show $int A = X - (\overline{X-A})$

Homework Equations

The Attempt at a Solution

[/B]
I am first trying to show that $int A \subseteq X - (\overline{X-A})$. Clearly if $x \in int A$, then $x$ will also be in $X$, because $int A \subseteq X$. All that remains is to show that $x \notin (\overline{X-A})$.

I could see a direct way of showing this inclusion, so I tried to prove it by contradiction, but that did not help either. Is there a direct way of showing this? Does anyone know if a direct proof is possible? Any hints?

 

[slider142]

Homework Statement

Let $(X,d)$ be some metric space, and let $A$ be some subset of the metric space. The interior of the set $A$, denoted as $int A$, is defined to be $\bigcup_{\alpha \in I} G_\alpha$, where $G_\alpha \subseteq A$ is open in $X$ for all $\alpha \in I$. The closure of $A$, denoted as $\overline{A}$, is defined to be $\bigcap_{\beta \in J} F_\beta$, where $A \subseteq F_\beta$ and $F_\beta$ is closed in $X$.

I am asked to show $int A = X - (\overline{X-A})$

Homework Equations

The Attempt at a Solution

[/B]
I am first trying to show that $int A \subseteq X - (\overline{X-A})$. Clearly if $x \in int A$, then $x$ will also be in $X$, because $int A \subseteq X$. All that remains is to show that $x \notin (\overline{X-A})$.

I could see a direct way of showing this inclusion, so I tried to prove it by contradiction, but that did not help either. Is there a direct way of showing this? Does anyone know if a direct proof is possible? Any hints?

A proof by contradiction is probably your best way forward in showing that something is *not* an element of a well defined set. Assume first that $x$ is in both $\text{int } A$ and $\overline{X-A}$. This means, with respect to your definition of closure, that $x$ must be an element of *every* closed set that contains $X - A$. Therefore, if we could find a closed set that contains $X - A$, but does not contain $x$, our contradiction would be complete.
$\text{int }A$ is an open subset of A, so it therefore has an empty intersection with $X - A$. What can we therefore say about the complement of $\text{int }A$ ? Do you see how this helps us in moving forward ?

 

[Bashyboy]

Yes, you hints have proved very helpful, so much so that I finished the proof. Now, though, I am having difficulty with the other inclusion. Here is what I have so far:

Suppose that $x \in X - (\overline{X-A})$. Then $x \in X$ but $x \notin (\overline{X-A}) = \bigcap_{j \in J} F_j$, which means $x \notin F_j$ for some $j \in J$. However, $x$ will be in $X-F_j$, which is open because $F_j$ is closed.

This is where I couldn't figure out how to proceed. I would like to show that $X-F_j \subseteq A$, as this would mean $X-F_j$ is apart of the union $\bigcup_{i \in I} G_i = int A$, which would conclude the proof. But I am uncertain of how to show this.