The Topology of Euclidean Space

The restriction $N:A\to\mathbb{R}$, $N(x)=\lVert x\rVert_2$, is continuous, because the ambient norm is continuous by [L1] and $A$ has the subspace topology.

Pseudocompactness gives that $N[A]$ is bounded. If $A\ne\varnothing$, choose an upper bound $M$ of $N[A]$; then $M\ge0$ because every norm is nonnegative.

If $A=\varnothing$ it is bounded. Otherwise every $x\in A$ satisfies $d_2(x,0)=\lVert x\rVert_2\le M<M+1$, 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 $p\in\overline A\setminus A$.

Define $f:A\to\mathbb{R}$ by $f(x):=1/\lVert x-p\rVert_2$. This is defined because $p\notin A$, so every denominator is positive.

For every real $M>0$, [L1] and [L2] give $x\in A$ with $\lVert x-p\rVert_2<1/M$; then $f(x)>M$. Hence $f[A]$ is unbounded.

The function $f$ is continuous on $A$: at $a\in A$ put $d:=\lVert a-p\rVert_2>0$. If $\lVert x-a\rVert_2<d/2$, then [L3] gives $\lVert x-p\rVert_2>d/2$, and $|f(x)-f(a)| \le \frac{2}{d^2}\lVert x-a\rVert_2.$ 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 $1\Leftrightarrow2$, $3\Rightarrow2\Rightarrow1$, and $1\Rightarrow4\Rightarrow3$ 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: $\overline B_2(c,r)$ lies in the ball of radius $r+1$ about $c$, and $S_2(c,r)\subseteq\overline B_2(c,r)$.

The complement of $\overline B_2(c,r)$ is open: if $\lVert x-c\rVert_2>r$, then the ball about $x$ of radius $(\lVert x-c\rVert_2-r)/2$ stays in the complement by [L2].

The complement of $S_2(c,r)$ is open: if $\lVert x-c\rVert_2\ne r$, then a ball about $x$ of radius $|\lVert x-c\rVert_2-r|/2$ avoids the sphere by [L2].

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

On $[t_{i-1},t_i]$ define $\gamma_i(t):=((t_i-t)/(t_i-t_{i-1}))v_{i-1}+((t-t_{i-1})/(t_i-t_{i-1}))v_i$. 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 $U=\varnothing$, 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 $\overline B_2(x,1)$ 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 $K_k:=\overline B_2(0,k+1)$. Every $K_k$ is compact by [L1].

If $x\in\mathbb R^n$, [L3] gives $k\ge1$ with $\lVert x\rVert_2<k$, 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 $q\mapsto(q,0,\ldots,0)$ 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 $x_i-r_i<a_i<x_i<b_i<x_i+r_i.$ 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 $z\in\mathbb R^n\setminus(\operatorname{span}\{x\}\cup\operatorname{span}\{y\})$. 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 $0<t\le1$ would give $z=-((1-t)/t)x\in\operatorname{span}\{x\}$, 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 $d:=\lVert a\rVert_2>0$. If $\lVert x-a\rVert_2<d/2$, then [L1] gives $\lVert x\rVert_2>d/2$.

For such $x$, $\lVert\rho(x)-\rho(a)\rVert_2\le\lVert x-a\rVert_2/\lVert x\rVert_2+\lVert a\rVert_2|1/\lVert x\rVert_2-1/\lVert a\rVert_2|\le4\lVert x-a\rVert_2/d$.

Step 1.2 gives the epsilon-delta condition at $a$, so $\rho$ is continuous on the punctured space. Also $\lVert\rho(x)\rVert_2=1$, so its image lies in $S^{n-1}$ and [L2] gives continuity with that codomain.

The scalar $c(x,t):=(1-t)+t/\lVert x\rVert_2$ is positive, since $1-t\ge0$, $t\ge0$, and $\lVert x\rVert_2>0$.

The coordinate projections on $P\times[0,1]$ are continuous by [L2]. By [L3] the map $x\mapsto\lVert x\rVert_2$ is continuous on $P$ and never $0$ there, so composing it with $u\mapsto 1/u$ gives a continuous $x\mapsto 1/\lVert x\rVert_2$ by [L4]; hence $c(x,t)=(1-t)+t/\lVert x\rVert_2$ 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 $H(x,1)=x/\lVert x\rVert_2$. If $s\in S^{n-1}$, then $\lVert s\rVert_2=1$ and $H(s,t)=s$.

Since $c(x,t)>0$ and $x\ne0$, $H(x,t)\ne0$. 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.