1. ## closure

Show that $X- E^{\circ} = \overline{X-E}$ where X is a metric space and E is a subset of X. Can we do this without using DMorgan's Laws (e.g. without using intersection of closed sets)?

Because you are taking away $E^{\circ} \subset E$ from $X$. Whereas $\overline{X-E} = X-E \cup (X-E)'$ where $(X-E)'$ is the set of limit points of $X-E$. So you are taking away more points from $X$. Adding back the limit points gives you $X- E^{\circ}$ (e.g. you are not adding any interior points).

2. First of all we should notice two facts:

1.
$X \setminus E^\circ \supset X \setminus E$

2.
$X \setminus \overline{X \setminus E} \subset (X \setminus X \setminus E) = E$

It follows from $E^\circ \subset E$

It follows from $X \setminus E \subset \overline{X \setminus E}$

Now, since $E^\circ$ is open, $X \setminus E^\circ$ is closed.
Since closure of A is the smallest closed set containing A from 1 we have:
$X \setminus E^\circ \supset \overline{X \setminus E}$

Now, since $\overline{X \setminus E}$ is closed, $X \setminus \overline{X \setminus E}$ is open.
Since interior of A is the largest open set in A we from 2 we have

$X \setminus \overline{X \setminus E} \subset E^\circ$

Thus: $\overline{X \setminus E} \supset X \setminus E^\circ$

So we have: $\overline{X \setminus E} = X \setminus E^\circ$