The Topology of Euclidean Space

The restriction $N:A\to \mathbb{R}$, $N(x)=\parallel x{\parallel }_2$, is continuous, because the ambient norm is continuous by [L1] and $A$ has the subspace topology.

Pseudocompactness gives that $N[A]$ is bounded. If $A\mathrm{\ne }\mathrm{\varnothing }$, choose an upper bound $M$ of $N[A]$; then $M\ge 0$ because every norm is nonnegative.

If $A=\mathrm{\varnothing }$ it is bounded. Otherwise every $x\in A$ satisfies $d_2(x,0)=\parallel x{\parallel }_2\le M<M+1$, so $A\subseteq B_{d_2}(0,M+1)$.

Thus $A$ is bounded in both cases.

Suppose, for contradiction, that $A$ is not closed. By [L1], fix $p\in \bar{A}\setminus A$.

Define $f:A\to \mathbb{R}$ by $f(x):=1\mathrm{/}\parallel x-p{\parallel }_2$. This is defined because $p\mathrm{\notin }A$, so every denominator is positive.

For every real $M>0$, [L1] and [L2] give $x\in A$ with $\parallel x-p{\parallel }_2<1\mathrm{/}M$; then $f(x)>M$. Hence $f[A]$ is unbounded.

The function $f$ is continuous on $A$: at $a\in A$ put $d:=\parallel a-p{\parallel }_2>0$. If $\parallel x-a{\parallel }_2<d\mathrm{/}2$, then [L3] gives $\parallel x-p{\parallel }_2>d\mathrm{/}2$, and $\mathrm{\mid }f(x)-f(a)\mathrm{\mid }\le \frac{2}{d^2}\parallel x-a{\parallel }_2\mathrm{.}$ Thus a sufficiently small Euclidean ball about $a$ maps into any prescribed real neighbourhood of $f(a)$.

Steps 3.1 and 2.2 contradict pseudocompactness through [L4]. Therefore $A$ is closed.

Conditions 1 and 2 are equivalent by [L1] and [L5].

Condition 3 implies condition 2 by [L2].

Suppose condition 1 holds. Every continuous $f:A\to \mathbb{R}$ then attains a maximum and a minimum by [L3], so condition 4 holds.

Suppose condition 4 holds. For every continuous $f:A\to \mathbb{R}$, its maximum and minimum bound $f[A]$, so $A$ is pseudocompact and condition 3 holds.

The implications $1\iff 2$, $3\Rightarrow 2\Rightarrow 1$, and $1\Rightarrow 4\Rightarrow 3$ prove all four conditions equivalent.

The ZF part of the chart is exactly [L1].

In the Euclidean metric, compactness is equivalent to each of countable compactness, limit point compactness, sequential compactness, and completeness together with total boundedness by [L2].

By [L3], this metric compactness is the compactness already occurring in step 1.1. Joining the two equivalence classes proves the asserted chart.

Both sets are bounded: ${\bar{B}}_2(c,r)$ lies in the ball of radius $r+1$ about $c$, and $S_2(c,r)\subseteq {\bar{B}}_2(c,r)$.

The complement of ${\bar{B}}_2(c,r)$ is open: if $\parallel x-c{\parallel }_2>r$, then the ball about $x$ of radius $(\parallel x-c{\parallel }_2-r)\mathrm{/}2$ stays in the complement by [L2].

The complement of $S_2(c,r)$ is open: if $\parallel x-c{\parallel }_2\mathrm{\ne }r$, then a ball about $x$ of radius $\mathrm{\mid }\parallel x-c{\parallel }_2-r\mathrm{\mid }\mathrm{/}2$ avoids the sphere by [L2].

Thus both sets are closed and bounded, hence compact by [L3].

On $[t_{i-1},t_i]$ define ${\gamma }_i(t):=((t_i-t)\mathrm{/}(t_i-t_{i-1}))v_{i-1}+((t-t_{i-1})\mathrm{/}(t_i-t_{i-1}))v_i$. Each coordinate is an affine real function of $t$, hence continuous.

At every shared endpoint $t_i$, the adjacent formulas both give $v_i$, so the pieces define one function $\gamma :[0,1]\to {\mathbb{R}}^n$.

Each interval $[t_{i-1},t_i]$ is closed in $[0,1]$, the finitely many intervals cover it, and each restriction of $\gamma$ is continuous by step 1.1. Thus $\gamma$ is continuous by [L1].

If the pieces lie in $A$, then $\gamma$ takes values in $A$. Its composite with the inclusion $A\hookrightarrow {\mathbb{R}}^n$ is the continuous map of step 3.1, so [L2] makes it continuous into the subspace $A$. Its endpoints are $v_0,v_m$, hence it is a path in $A$.

Let $y\in R_a$. Choose $r>0$ with $B(y,r)\subseteq U$. For every $z\in B(y,r)$, the segment from $y$ to $z$ lies in $B(y,r)$ by [L2].

Let $y\in U\setminus R_a$ and choose $r>0$ with $B(y,r)\subseteq U$. If some $z\in B(y,r)$ lay in $R_a$, a path from $a$ to $z$ followed by the segment from $z$ to $y$ would put $y$ in $R_a$, a contradiction.

Concatenating a polygonal path from $a$ to $y$ with that segment gives a polygonal path from $a$ to $z$ in $U$. Thus $B(y,r)\subseteq R_a$, so $R_a$ is open in $U$.

Hence $B(y,r)\subseteq U\setminus R_a$, so the complement is open in $U$. Therefore $R_a$ is clopen in $U$.

Suppose $U$ is connected. If $U=\mathrm{\varnothing }$, then polygonal connectedness, path-connectedness, and connectedness all hold vacuously. Otherwise choose $a\in U$. The reachable set $R_a$ is nonempty and clopen by [L1], so [L3] gives $R_a=U$.

Polygonal connectedness implies path-connectedness, and path-connectedness implies connectedness, by [L2].

In the nonempty case, every point of $U=R_a$ is joined to $a$ by a polygonal path; reversing one such path and concatenating it with another joins any two points of $U$. Together with the empty case, connectedness implies polygonal connectedness.

Steps 2.1 and 1.2 give all three equivalences.

Let $x\in C$. Since $U$ is open, choose $r>0$ with $B(x,r)\subseteq U$. The ball is connected by [L1], meets $C$ at $x$, and so lies in $C$ by maximality of the component.

Therefore every point of $C$ has a Euclidean ball contained in $C$, so $C$ is open in ${\mathbb{R}}^n$.

The component $C$ is connected and now open, so [L3] makes it polygonally connected.

For $x,y\in {\mathbb{R}}^n$, the one-segment path $t\mapsto (1-t)x+ty$ joins them. Hence ${\mathbb{R}}^n$ is polygonally connected.

Let $x\in {\mathbb{R}}^n$ and let $U$ be open with $x\in U$. Choose $r>0$ with $B(x,r)\subseteq U$. Each pair of points in this ball is joined by its segment, which stays in the ball by [L4].

It is path-connected and therefore connected by [L3].

Thus every open neighbourhood contains an open path-connected ball, so ${\mathbb{R}}^n$ is locally path-connected, and it is locally connected by [L3].

For $x\in {\mathbb{R}}^n$, the closed ball ${\bar{B}}_2(x,1)$ is compact by [L1] and contains the open ball $B_2(x,1)$ about $x$. Thus $x$ has a compact neighbourhood.

For each $k\in \mathbb{N}$ put $K_k:={\bar{B}}_2(0,k+1)$. Every $K_k$ is compact by [L1].

If $x\in {\mathbb{R}}^n$, [L3] gives $k\ge 1$ with $\parallel x{\parallel }_2<k$, so $x\in K_k$. Hence ${\mathbb{R}}^n={\bigcup }_{k\in \mathbb{N}}{K}_k$.

Step 1.1 gives local compactness and step 2.1 gives $\sigma$-compactness.

For $m=0$, $C^0$ is a one-point set by [L2], hence at most countable by [L3].

Assume $C^m$ is at most countable.

The product $C^{m+1}$ is canonically $C^m\times C$, so it is at most countable by [L1].

By induction, $C^m$ is at most countable for every $m\in \mathbb{N}$.

By [L1], ${\mathbb{Q}}^n$ is at most countable. It is infinite because the injection $q\mapsto (q,0,\dots ,0)$ embeds $\mathbb{Q}$ in it, hence it is countable.

Let $U={\prod }_{i<n}{U}_i$ be a nonempty basic product-open set. Every $U_i$ is nonempty and open, so [L2] gives a rational $q_i\in U_i$; finite choice supplies the tuple $q=(q_i{)}_{i<n}\in {\mathbb{Q}}^n\cap U$.

Let $x\in U={\prod }_{i<n}{U}_i$ be a basic product-open neighbourhood. For each $i<n$, use [L6] to choose $r_i>0$ with $(x_i-r_i,x_i+r_i)\subseteq U_i$, then use [L2] to choose rationals $x_i-r_i<a_i<x_i<b_i<x_i+r_i\mathrm{.}$ Finite choice assembles these choices, and then $x\in {\prod }_{i<n}(a_i,b_i)\subseteq U$.

Every nonempty open subset contains a nonempty basic product-open set about each of its points by [L3], so step 1.2 shows that every nonempty open subset meets ${\mathbb{Q}}^n$. Thus ${\mathbb{Q}}^n$ is dense.

Step 1.3 and [L4] show that rational open boxes form a basis. They are indexed by a subset of ${\mathbb{Q}}^{2n}$, which is at most countable by [L1], so the basis is at most countable by [L5].

Choose $z\in {\mathbb{R}}^n\setminus (\mathrm{span}\{x\}\cup \mathrm{span}\{y\})$. Such a vector exists: if the two spans differ, $x+y$ lies in neither; if they agree, [L1] gives a vector outside their common proper subspace.

The segment from $x$ to $z$ avoids $0$: an equality $(1-t)x+tz=0$ with $0<t\le 1$ would give $z=-((1-t)\mathrm{/}t)x\in \mathrm{span}\{x\}$, contrary to step 1.1. The segment from $z$ to $y$ similarly avoids $0$.

The two segments therefore form a polygonal path in ${\mathbb{R}}^n\setminus \{0\}$ from $x$ to $y$ by [L3].

Since $x,y$ were arbitrary nonzero vectors, the punctured space is polygonally connected.

Put $d:=\parallel a{\parallel }_2>0$. If $\parallel x-a{\parallel }_2<d\mathrm{/}2$, then [L1] gives $\parallel x{\parallel }_2>d\mathrm{/}2$.

For such $x$, $\parallel \rho (x)-\rho (a){\parallel }_2\le \parallel x-a{\parallel }_2\mathrm{/}\parallel x{\parallel }_2+\parallel a{\parallel }_2\mathrm{\mid }1\mathrm{/}\parallel x{\parallel }_2-1\mathrm{/}\parallel a{\parallel }_2\mathrm{\mid }\le 4\parallel x-a{\parallel }_2\mathrm{/}d$.

Step 1.2 gives the epsilon-delta condition at $a$, so $\rho$ is continuous on the punctured space. Also $\parallel \rho (x){\parallel }_2=1$, so its image lies in $S^{n-1}$ and [L2] gives continuity with that codomain.

The scalar $c(x,t):=(1-t)+t\mathrm{/}\parallel x{\parallel }_2$ is positive, since $1-t\ge 0$, $t\ge 0$, and $\parallel x{\parallel }_2>0$.

The coordinate projections on $P\times [0,1]$ are continuous by [L2]. By [L3] the map $x\mapsto \parallel x{\parallel }_2$ is continuous on $P$ and never $0$ there, so composing it with $u\mapsto 1\mathrm{/}u$ gives a continuous $x\mapsto 1\mathrm{/}\parallel x{\parallel }_2$ by [L4]; hence $c(x,t)=(1-t)+t\mathrm{/}\parallel x{\parallel }_2$ is continuous, being built from continuous functions by sums, scalar multiples and products as in [L4], and each coordinate $H_i(x,t)=c(x,t)x_i$ is continuous. Componentwise continuity in [L4] makes $H$ continuous as a map into ${\mathbb{R}}^n$.

Substituting $t=0$ and $t=1$ gives $H(x,0)=x$ and $H(x,1)=x\mathrm{/}\parallel x{\parallel }_2$. If $s\in S^{n-1}$, then $\parallel s{\parallel }_2=1$ and $H(s,t)=s$.

Since $c(x,t)>0$ and $x\mathrm{\ne }0$, $H(x,t)\mathrm{\ne }0$. Hence $H$ takes values in $P$, and [L5] makes it continuous as a map $P\times [0,1]\to P$.

These identities and step 2.1 prove the statement.

Choose a path $\gamma :[0,1]\to {\mathbb{R}}^n\setminus \{0\}$ from $x$ to $y$ as in [L1], and define $\eta :=\rho \circ \gamma$, where $\rho (z)=z\mathrm{/}\parallel z{\parallel }_2$.

The map $\eta$ is continuous and takes values in $S^{n-1}$ by [L2]. Since $\rho (x)=x$ and $\rho (y)=y$, it joins $x$ to $y$ in the sphere.

Thus the sphere is path-connected, and it is connected by [L3].

Suppose $h:\mathbb{R}\to {\mathbb{R}}^n$ is a homeomorphism, and put $a:=h^{-1}(0)$.

Restricting $h^{-1}$ to ${\mathbb{R}}^n\setminus \{0\}$ gives a continuous surjection onto $\mathbb{R}\setminus \{a\}$. The source is connected by [L1], so the target is connected by [L2].

Choose $a-1<a<a+1$. Both endpoints lie in $\mathbb{R}\setminus \{a\}$, but $a$ does not, so this subset is not order-convex and therefore not connected by [L3].

Steps 1.2 and 1.3 contradict one another. Thus no such homeomorphism exists.