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

If $\sigma$ and $\sigma'$ are metrics on one set $Z$, then $\mathcal{T}_\sigma = \mathcal{T}_{\sigma'}$ holds exactly when both of the following do: for every $z \in Z$ and real $\varepsilon > 0$ there is a real $\delta > 0$ with $\sigma'(z,w) < \varepsilon$ whenever $\sigma(z,w) < \delta$, and the same with $\sigma$ and $\sigma'$ interchanged. Indeed the two conditions say that the two identity maps are $\varepsilon$-$\delta$ continuous, which by [L1] says that each topology is contained in the other.

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

For claim 3 put $\sigma(x,y) := |1/x - 1/y|$ 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 $1/x = 1/y$ 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 $\varepsilon > 0$ be real; put $\delta := \min\{\, a/2,\ \varepsilon/(1 + 2/a^2) \,\}$, a positive real. For $y \in P$ with $|a - y| < \delta$ one has $y > a - a/2 = a/2$, hence $ay > a^2/2$ and $\sigma(a,y) = |a-y|/(ay) < 2|a-y|/a^2$, so $\rho_P(a,y) < |a-y| \cdot (1 + 2/a^2) < \varepsilon$.

The sequence $x_k := 1/(k+2)$ 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 $(\mathbb{R}, |x-y|)$.

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 $\rho'$-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 $\rho'(y_k, h(x)) = \rho(h^{-1}(y_k), x) \to 0$, that is $y_k \to h(x)$ in $(Y,\rho')$. So $(Y,\rho')$ 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 $\rho'$ itself as subspace metric, a set $U \subseteq X$ is $\rho$-open exactly when $h[U]$ is $\rho'$-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 $\mathcal{T}_{\rho'} = \mathcal{T}_e$.

Claim 2, topology: apply step 1.1 to $\rho$ and $d$ on $X$, which is legitimate by [A1], and restrict the two resulting $\varepsilon$-$\delta$ 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 $\delta := \varepsilon$, 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 $(\mathbb{R}, |x-y|)$, 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 $(1/x_k)$ are Cauchy sequences of reals, so by [L6] they converge, say $x_k \to L$ and $1/x_k \to c$; and $L \ge 0$ and $c \ge 0$, all terms being positive.

Claim 1 is established: $\rho'$ is a complete metric on $Y$ with $\mathcal{T}_{\rho'} = \mathcal{T}_e$, 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: $x_k \cdot (1/x_k) = 1$ for every $k$, so $L c = 1$ by multiplicativity of limits; hence $L \ne 0$, so $L > 0$ and $L \in P$, and $c = 1/L$.

Hence $\rho_P(x_k, L) = |x_k - L| + |1/x_k - 1/L| \to 0$ 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.