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.

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 $N_r(p)\bigcup E\setminus \{p\}$ and call it $s.$ Now, we can make a new neighborhood $N_s(p),$ which contains none of the points in $N_r(p)\bigcup E\setminus \{p\}$ 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 $E\setminus \{p\}$ 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.)

(b) Suppose $E$ is closed. Then it contains its limit points, so $E=E\cup E^{\prime }=\bar{E}.$ Conversely, suppose $E=\bar{E}.$ By (a), $E$ is closed.

(c) Suppose that $E\subset F\subset X,$ and that $F$ is closed. Suppose $p\in \bar{E}.$ If $p\in E,$ then $p\in F$ because $E\subset F.$ If $p\in E^{\prime },$ then it must be in $F$ also, since $F$ contains all points of $E,$ and is closed, and thus must contain the limit points of $E.$

If $E$ is closed, then we are done, because a set equals its closure if it is closed.

Suppose $p$ is a limit point of $E.$ Then every neighborhood of $p$ contains some $q\in E,q\ne p.$ Since $q\in E,q\in \bar{E}$, so $p$ is a limit point of $\bar{E}.$

Conversely, suppose $p$ is a limit point of $\bar{E}.$ Then, every neighborhood $N$ of $p$ contains a point of $q\in \bar{E},q\ne p.$ If $q\in E,$ then $N$ clearly contains a point in $E.$ Otherwise, $q\in E^{\prime },q\mathrm{i̸}nE.$ Now, since $q\in N$ and every neighborhood is an open set, $q$ has some neighborhood $M\subset N.$ Since $q\in E^{\prime },$ $M$ contains some point $s\in E.$ Since $M\subset N,$ $s\in N,$ and therefore $N$ contains a point in $E.$ Thus, all neighborhoods of $p$ contain some point in $E,$ and $p$ is a limit point of $E.$

Consider $P=\{\frac{1}{n}|n\in \mathbb{N}\}.$ Then $P$ has one limit point, $0,$ but $P^{\prime }=\{0\}$ has no limit points, since the only number that contains $0$ in all its neighborhoods is $0$ itself (see From this, it's evident that a finite....)

Let $p\in \bigcup _{i=1}^n\overline{A_i}.$ Then, $p\in \overline{A_i}$ for some $i.$ If $p\in A_i,$ then $p\in \bigcup _{i=i}^n{A}_i=B_n,$ so $p\in \overline{B_n}.$ If $p$ is in some $A_i^{\prime },$ then every neighborhood of $p$ contains some point $q\in A_i,q\ne p,$ and since $A_i\subset B_n,$ $q\in B_n,$ so $p\in \overline{B_n}.$

Conversely, let $p\in \overline{B_n}.$ If $p\in B_n,$ $p\in A_i$ for some $i,$ and thus $p\in \overline{A_i}\subset \bigcup _{i=1}^n\overline{A_i}.$ If $p$ is only in $B_n^{\prime },$ suppose for contradiction that $p\notin \bigcup _{i=1}^n\overline{A_i}.$ For each $i,$ let $N_i$ be a neighborhood centered at $p$ with $N_i\cap A_i=\mathrm{\varnothing }.$ Then, $N=\bigcap _{i=1}^n{N}_i$ is a neighborhood of $p$ (see (a) - For any collection $\{G_a\}$ of...,) but $N\cap N_i=\mathrm{\varnothing }$ for all $i,$ so $N\cap B_n=\mathrm{\varnothing }.$ But, this contradicts our hypothesis that $p\in \overline{B_n},$ so our contradictory assumption must be invalid, and $p\in \bigcup _{i=1}^n\overline{A_i}.$

Suppose $y\in E.$ Then $y\in \bar{E}.$ Suppose $y\notin E.$ Now, by hypothesis, for every there is some $x$ such that because otherwise, $x$ would be an upper bound on $E.$ Therefore, every neighborhood $N_h(y)$ contains some $x\in E,$ and thus $y$ is a limit point of $E$ and $y\in \bar{E}.$

Suppose $E$ is open relative to $Y.$ Then, for every $p\in E,$ there is some such that $q\in X$ implies that $q\in Y.$ Let $V_p$ be the set of all $q\in X$ where (and thus $q\in Y$) and let

$$G=\bigcup _{p\in E}{V}_p.$$

Then, since each $V_p$ is an open subset of $X,$ so is $G.$ Now, since $p\in V_p$ for each $p\in E,$ $E\subset G\cap Y.$ Also, since $V_p\cap Y\subset E$ for every $p\in E,$ $G\cap Y\subset E,$ and $E=G\cap Y.$

Conversely, suppose $E=Y\cap G$ for some open subset $G$ of $X.$ Now, suppose $p\in E.$ Then, $p\in G,$ and there is some neighborhood $V_p\subset G.$ Then, $V_p\cap Y\subset E,$ so $E$ is open relative to $Y.$