First, for $n=0$, ${\mathbb{R}}^0$ is just the empty set, so the metric axioms are vacuously satisfied for all points in the set. Now, for $n\ge 1,$

• Let $p,q\in R^n,p\ne q.$ Then so

• Let $p\in R^n.$ Now, $p-p=0,$ so $|p-p|=0,$ so $d(p,p)=0.$

• Let $p,q\in R^n.$ $|p-q|=|q-p|,$ so $d(p,q)=d(q,p).$

• Let $p,q,r\in R^n.$ $|p-q|\le |p-r|+|r-p|,$ so $d(p,q)\le d(p,r)+d(r,p).$

Therefore, $d$ is a metric on $R^n$ and $(R^n,d)$ is a metric space.

Let $\mathbf{y},\mathbf{z}$ be points in a ball with center $\mathbf{x}$ and radius $r$. Then, by definition, and Suppose We will show that

So, $\mathbf{p}$ is within our ball and therefore all balls are convex.

TODO

Suppose $N_r(p)$ is a neighborhood in $X.$ Let $q\in N_r(p).$ We need to show that $q$ is an interior point of $N_r(p).$ Let $s=r-d(p,q);$ because , we have Now let $N_s(q)$ be the neighborhood of radius $s$ around $q.$ We need to show that $N_s(q)\subset N_r(p).$ Suppose $x\in N_s(q).$ First note that because $s=r-d(p,q),$ $d(p,q)=r-s.$ Now,

Therefore, $N_s(q)\subset N_r(p),$ so $q$ is an interior point of $N_r(p),$ and since $q$ was arbitrary, every point of $N_r(p)$ is interior. Hence, $N_r(p)$ is open.

Let $p$ be a limit point of $E$ and let $N_r(p)$ be a neighborhood of $p.$ Suppose that $N_r(p)$ contains only finitely many points of $E.$ Since we have finitely many points, we can inspect each and find the minimum distance from $p$ to any point in $N_r(p)\bigcup E\setminus \left\{p\right\}$ and call it $s.$ Now, we can make a new neighborhood $N_s(p),$ which contains none of the points in $N_r(p)\bigcup E\setminus \left\{p\right\}$ since they're all at least $s$ away from $p,$ by construction. But then, $p$ is not a limit point of $E,$ since it has a neighborhood that contains no points of $E\setminus \left\{p\right\}$ Therefore, we have a contradiction, and $N_r(p)$ must therefore contain infinitely many points.

Suppose $x\in {\left(\bigcup _{\alpha }{E}_{\alpha }\right)}^c.$ Then, $x\notin \bigcup _{\alpha }{E}_{\alpha },$ so $x$ is not in any $E_{\alpha }.$ Therefore, for every $E_{\alpha },$ $x\in E_{\alpha }^c,$ and thus $x\in \bigcap _{\alpha }\left(E_{\alpha }^c\right).$ Conversely, suppose $x\in \bigcap _{\alpha }\left(E_{\alpha }^c\right).$ Then, $x$ is in every $E_{\alpha }^c,$ that is, $x$ is not in any $E_{\alpha }.$ Therefore, $x\in {\left(\bigcup _{\alpha }{E}_{\alpha }\right)}^c.$

First, consider the case that $E^c$ is empty, and therefore open. If $E^c$ has no limit points, it is vacuously closed. Suppose $E^c$ has a limit point $x.$ Since $E$ is empty, $x$ must be in $E^c,$ therefore $x$ is closed. Now, consider the case that $E^c$ is empty, and therefore closed. If $E$ is empty, it is open, and the theorem is satisfied. If $E$ is not empty, a point $x$ in $E$ has only points in $E$ in any neighborhood, since all points are in $E,$ and therefore $E$ is open.

Now we deal with the cases where neither $E$ nor $E^c$ are empty.

Now, let $E^c$ be closed. Let $x\in E.$ Since $E^c$ is closed, $x$ is not a limit point of $E^c,$ that is $x$ has some neighborhood that doesn't contain a point in $E^c$ and must therefore be a subset of $E.$ Therefore, $x$ is an interior point of $E,$ and $E$ is open.

Conversely, assume $E$ is open. Let $x$ be a limit point of $E^c.$ Suppose, for the sake of contradiction, that $x\in E.$ Then, since $E$ is open, $x$ is an interior point of $E$ and has some neighborhood that is a subset of $E.$ This is a contradiction, since every neighborhood of $x$ must contain at least one point of $E^c$ to be a limit point of $E^c.$ Therefore, $x$ must be in $E^c,$ and it follows that $E^c$ is closed.

Let $G=\bigcup _{\alpha }{G}_{\alpha },x\in G.$ Then $x$ is in some $G_{\alpha }$ for some $\alpha ,$ and is an interior point of that $G_{\alpha },$ since $G_{\alpha }$ is open. Therefore, $x$ has some neighborhood that is a subset of $G_{\alpha }$ and therefore of $G,$ so $x$ is an interior point of $G$ and $G$ is open - this shows (a).

Note that

$$\begin{array}{cc}(\bigcap _{\alpha }{F}_{\alpha }{)}^c=\bigcup _{\alpha }{F}_{\alpha }^c,& \text{(e)}\end{array}$$

and by (a) above, (e) is open. Then its complement, $\bigcap \alpha F_{\alpha }$ is closed, and we've shown (b).

Now, let $x$ be in $G=\bigcap _{i=1}^n{G}_i,$ so $x$ is in every $G_i,$ and has a neighborhood $N_i$ in every $G_i$ with radius Let $r$ be $min\{r_1,\dots ,r_n\}.$ Then, $x$ has a neighborhood $N$ of radius $r$ in every $G_i,$ and thus in $G,$ so $x$ is an interior point of $G,$ and we've shown (c).

Now, $(\bigcup _{i=1}^n{F}_i{)}^c=\bigcap _{i=1}^n{F}_i^c$ is open by (c), so its complement, $\bigcup _{i=1}^n{F}_i$ is closed, and we've shown (d).

Let $p\in X,p\notin E^{\prime }.$ Then some neighborhood $N$ of $p$ contains no points in $E,$ other than possibly $p$ itself. If $N$ contains only $p,$ then $p$ is not a limit point of $E^{\prime }.$ Suppose, for the sake of contradiction, some point $q\in N,q\ne p$ is a limit point of $E^{\prime }.$ Then, every neighborhood of $q$ contains some point $m$ in $E^{\prime }.$ Let $M\subset N,p\notin M,$ be such a neighborhood, and let $m\in M,m\ne q$ be a limit point of $E$. Now, $m$ has neighborhoods wholly in $N,$ and such neighborhoods can have no point in $E,$ so we have a contradiction, and therefore $q$ is not a limit point of $E^{\prime }.$ Hence, $p$ is an interior point of $E^{\prime c},$ and $E^{\prime c}$ is open, and therefore $E^{\prime }$ is closed.

(a) Suppose $p\in X$ and $p\notin \bar{E}.$ Then $p$ is not in $E$ and is not in $E^{\prime },$ and is in fact in ${\bar{E}}^c.$ Now, since $p$ is not a limit point of $E,$ it has some neighborhood $N_r(p)$ that does not intersect $E.$ Any point $x$ in $N_r(p)$ is an interior point of $N_r(p)$, and therefore has its own neighborhood $N_{ϵ}(x)$ that does not intersect $E,$ and therefore $x$ is not a limit point of $E.$ Thus, any point in ${\bar{E}}^c$ is an interior point of ${\bar{E}}^c,$ and ${\bar{E}}^c$ is therefore open and its complement, $\bar{E},$ is closed (since a set is open iff its complement is closed.)