Complete metrizability: admitting a topologically equivalent complete metric is preserved by homeomorphism and by closed subspaces, and $(0,\mathrm{\infty })$ has it without being complete

If $\sigma$ and ${\sigma }^{\mathrm{\prime }}$ are metrics on one set $Z$, then ${\mathcal{T}}_{\sigma }={\mathcal{T}}_{{\sigma }^{\mathrm{\prime }}}$ holds exactly when both of the following do: for every $z\in Z$ and real $\epsilon >0$ there is a real $\delta >0$ with ${\sigma }^{\mathrm{\prime }}(z,w)<\epsilon$ whenever $\sigma (z,w)<\delta$, and the same with $\sigma$ and ${\sigma }^{\mathrm{\prime }}$ interchanged. Indeed the two conditions say that the two identity maps are $\epsilon$-$\delta$ continuous, which by [L1] says that each topology is contained in the other.

For claim 1 let $\rho$ be as in [A1] and put ${\rho }^{\mathrm{\prime }}(y,y^{\mathrm{\prime }}):=\rho (h^{-1}(y),h^{-1}(y^{\mathrm{\prime }}))$ for $y,y^{\mathrm{\prime }}\in Y$. This is a metric on $Y$: (M2) and (M3) are inherited pointwise from $\rho$, and (M1) holds because $h^{-1}$ is injective, so ${\rho }^{\mathrm{\prime }}(y,y^{\mathrm{\prime }})=0$ gives $h^{-1}(y)=h^{-1}(y^{\mathrm{\prime }})$ and hence $y=y^{\mathrm{\prime }}$. By construction ${\rho }^{\mathrm{\prime }}(h(x),h(x^{\mathrm{\prime }}))=\rho (x,x^{\mathrm{\prime }})$, so $h:(X,\rho )\to (Y,{\rho }^{\mathrm{\prime }})$ is a bijective isometric embedding.

For claim 3 put $\sigma (x,y):=\mathrm{\mid }1\mathrm{/}x-1\mathrm{/}y\mathrm{\mid }$ for $x,y\in P$. This is a metric on $P$: it is nonnegative and symmetric, it satisfies the triangle inequality because the absolute value does, and $\sigma (x,y)=0$ gives $1\mathrm{/}x=1\mathrm{/}y$ and hence $x=y$.

A sum of two metrics on one set is again a metric, since symmetry and the triangle inequality add, the sum of two nonnegative reals is nonnegative, and the sum vanishes exactly when both summands do. Hence ${\rho }_P=d+\sigma$ is a metric on $P$, and $d(x,y)\le {\rho }_P(x,y)$ and $\sigma (x,y)\le {\rho }_P(x,y)$ for all $x,y\in P$.

Let $a\in P$ and let $\epsilon >0$ be real; put $\delta :=\min \{a\mathrm{/}2,\epsilon \mathrm{/}(1+2\mathrm{/}{a}^2)\}$, a positive real. For $y\in P$ with $\mathrm{\mid }a-y\mathrm{\mid }<\delta$ one has $y>a-a\mathrm{/}2=a\mathrm{/}2$, hence $ay>a^2\mathrm{/}2$ and $\sigma (a,y)=\mathrm{\mid }a-y\mathrm{\mid }\mathrm{/}(ay)<2\mathrm{\mid }a-y\mathrm{\mid }\mathrm{/}{a}^2$, so ${\rho }_P(a,y)<\mathrm{\mid }a-y\mathrm{\mid }\cdot (1+2\mathrm{/}{a}^2)<\epsilon$.

The sequence $x_k:=1\mathrm{/}(k+2)$ has all its terms in $P$ and converges in $\mathbb{R}$ to $0$, which is not in $P$; so $P$ is not sequentially closed in $\mathbb{R}$ and therefore not closed in $(\mathbb{R},\mathrm{\mid }x-y\mathrm{\mid })$.

For claim 2 let $\rho$ be as in [A1]. Since ${\mathcal{T}}_{\rho }={\mathcal{T}}_d$ and $A$ is closed in $(X,d)$, the set $A$ is closed in $(X,\rho )$ as well, so $(A,{\rho }_A)$ is complete by [L3], ${\rho }_A$ being the restriction of $\rho$ to $A\times A$.

Claim 1, completeness: let $(y_k)$ be a ${\rho }^{\mathrm{\prime }}$-Cauchy sequence in $Y$. By step 1.2 the sequence $(h^{-1}(y_k))$ is $\rho$-Cauchy, so by [A1] it converges in $(X,\rho )$ to some $x\in X$, and then ${\rho }^{\mathrm{\prime }}(y_k,h(x))=\rho (h^{-1}(y_k),x)\to 0$, that is $y_k\to h(x)$ in $(Y,{\rho }^{\mathrm{\prime }})$. So $(Y,{\rho }^{\mathrm{\prime }})$ is complete.

Claim 1, topology: by [L4] applied to the bijective isometric embedding $h$ of step 1.2, whose image is all of $Y$ with ${\rho }^{\mathrm{\prime }}$ itself as subspace metric, a set $U\subseteq X$ is $\rho$-open exactly when $h[U]$ is ${\rho }^{\mathrm{\prime }}$-open. And $U$ is $d$-open exactly when $h[U]$ is $e$-open, since $h[U]=(h^{-1}{)}^{-1}[U]$ is $e$-open for $d$-open $U$ by continuity of $h^{-1}$, and conversely $U=h^{-1}[h[U]]$ is $d$-open for $e$-open $h[U]$ by continuity of $h$. As ${\mathcal{T}}_{\rho }={\mathcal{T}}_d$ by [A1], the two equivalences give ${\mathcal{T}}_{{\rho }^{\mathrm{\prime }}}={\mathcal{T}}_e$.

Claim 2, topology: apply step 1.1 to $\rho$ and $d$ on $X$, which is legitimate by [A1], and restrict the two resulting $\epsilon$-$\delta$ conditions to points of $A$; since ${\rho }_A$ and $d_A$ are the restrictions of $\rho$ and $d$, the same $\delta$s witness the two conditions of step 1.1 for ${\rho }_A$ and $d_A$ on $A$, whence ${\mathcal{T}}_{{\rho }_A}={\mathcal{T}}_{d_A}$.

Claim 3, topology: by step 1.4 the identity $(P,{\rho }_P)\to (P,d)$ satisfies the first condition of step 1.1 with $\delta :=\epsilon$, and by step 1.5 the identity $(P,d)\to (P,{\rho }_P)$ satisfies the second; so ${\mathcal{T}}_{{\rho }_P}={\mathcal{T}}_d$.

Claim 3, failure of completeness for $d$: were $(P,d)$ complete, [L3] would make $P$ closed in $(\mathbb{R},\mathrm{\mid }x-y\mathrm{\mid })$, contradicting step 1.6. So $(P,d)$ is not complete.

Claim 3, completeness of ${\rho }_P$: let $(x_k)$ be a ${\rho }_P$-Cauchy sequence in $P$. By the two inequalities of step 1.4 both $(x_k)$ and $(1\mathrm{/}{x}_k)$ are Cauchy sequences of reals, so by [L6] they converge, say $x_k\to L$ and $1\mathrm{/}{x}_k\to c$; and $L\ge 0$ and $c\ge 0$, all terms being positive.

Claim 1 is established: ${\rho }^{\mathrm{\prime }}$ is a complete metric on $Y$ with ${\mathcal{T}}_{{\rho }^{\mathrm{\prime }}}={\mathcal{T}}_e$, so ${\mathcal{T}}_e$ is completely metrizable.

Claim 2 is established: ${\rho }_A$ is a complete metric on $A$ with ${\mathcal{T}}_{{\rho }_A}={\mathcal{T}}_{d_A}$, so ${\mathcal{T}}_{d_A}$ is completely metrizable.

Continuing step 2.6: $x_k\cdot (1\mathrm{/}{x}_k)=1$ for every $k$, so $Lc=1$ by multiplicativity of limits; hence $L\mathrm{\ne }0$, so $L>0$ and $L\in P$, and $c=1\mathrm{/}L$.

Hence ${\rho }_P(x_k,L)=\mathrm{\mid }{x}_k-L\mathrm{\mid }+\mathrm{\mid }1\mathrm{/}{x}_k-1\mathrm{/}L\mathrm{\mid }\to 0$ by additivity of limits, that is $x_k\to L$ in $(P,{\rho }_P)$ with $L\in P$; every ${\rho }_P$-Cauchy sequence in $P$ therefore converges in $P$, and $(P,{\rho }_P)$ is complete.

Claim 3 is established by step 2.4, step 2.5 and step 4.1, and claims 1 and 2 by step 3.1 and step 3.2.