Metric spaces and openness I am asked to prove two things. I would like to know if the proof was elaborate and concise.

I would also like to know if proving reductio ad absurdum is looked down upon. I have heard from my professor say it's not recommended, but my analysis book says it is an invaluable tool.

a) Prove that in a metric space the complement of a point is open.

Suppose that the point $x$ is contained in a closed interval [x]. Now suppose for contradiction the complement of the point is closed, that is, $a&lt;=x$ and $x&lt;= b$. Now, the absurdity is at hand: $[x] \cup [a, b] \leftrightarrow [a, b]$ when they were originally meant to be disjoint. So the complement of the point must be open, that is, $a &lt; x &lt; b$.

b) Prove that any set in a metric space is an intersection of open sets. 

No clue here. I would gladly appreciate any clues. I am self-learning this.
Let $(X,d)$ be a metric space.

Let $x \in X$.

We wish to show that $X\setminus \{x\}$ is open.

By definition, this means showing that for each $y \in X \setminus \{x\}$ there exists an $\epsilon\in(0,\infty)$ such that &hellip;.

Fix a specific $y$. Then the distance between $x$ and $y$ is $d(x,y)$.

Let $\epsilon = \text{something or other to do with }d(x,y)$.

Then &hellip;
a) Prove that in a metric space the complement of a point is open.

Suppose there exists a metric space $x\in X$. We wish to show $X - {x}$ is open. Thus, define

$$\forall y \in M : D(x, y) &lt; \epsilon$$

By definition, $ \forall y \in M, S_{\epsilon}(x)$ is an open ball.  Because open balls are open, it has been shown the complement of a point is open.

b) Prove that any set in a metric space is an intersection of open sets. 

Suppose there exists a set $E \subset M, E = E_{1} \cup E_{n}$. Next, suppose $e \in E$. Then $e \in E_{n} \forall n$. Since $E_{n}$ is open we have $B_{r_{i}}(e) \subset E_{n}$. If we take $r = min (r_1, r_2, ... r_n)$, then $B_{r}(e) \subset E$. QED.