Metrization: Urysohn, Nagata–Smirnov, Bing, Smirnov: Examples and Counterexamples

The zero-one function $d(x,y)=0$ for $x=y$ and $d(x,y)=1$ otherwise is a metric inducing the discrete topology; singleton sets form a discrete basis.

Any basis must contain $\{x\}$ for every $x$: applying the basis condition to the open set $\{x\}$ produces a basis member containing $x$ and contained in $\{x\}$. Thus a countable basis would inject the uncountable set $\mathbb{R}$ into a countable set, contradicting [L1].

This gives the claimed profile.

By [L1] the space is regular, and by [L2] the countable set $\mathbb{Q}$ is dense, so it is separable.

Suppose $(B_n{)}_{n\in \mathbb{N}}$ is a basis. For each real $x$, the basis condition for $[x,x+1)$ yields a least-index $B_{m(x)}$ with $x\in B_{m(x)}\subseteq [x,x+1)$.

If $m(x)=m(y)$ and $x<y$, then $x\in B_{m(y)}\subseteq [y,y+1)$, impossible. Thus $x\mapsto m(x)$ injects $\mathbb{R}$ into $\mathbb{N}$, contradicting $\mathbb{R}$ is uncountable (Cantor's nested intervals, 1874).

Therefore the lower-limit line is not second countable. If it were metrizable, its separability from step 1.1 and [L3] would make it second countable, another contradiction.

Suppose the displayed assertion is true. By [L1], the lower-limit line would be metrizable.

The direct basis argument assigns to each $x$ the least member of a putative countable basis contained in $[x,x+1)$ and containing $x$; equality of assigned members forces equality of their left endpoints. Thus no countable basis exists, since it would inject $\mathbb{R}$ into $\mathbb{N}$, contrary to [L2].

The rational-density argument makes the line separable, so metrizability from step 1.1 would imply second countability by Assuming countable choice, a metrizable space is second countable if and only if it is separable if and only if it is Lindelöf, contradicting step 1.2.

Hence the regular lower-limit line refutes the displayed assertion.

Tangent disks and ordinary disks satisfy [L1]: intersections at a positive-height point contain a small ordinary disk, and a tangent disk at a boundary point contains a smaller tangent disk. Distinct points have disjoint such members: use small ordinary disks above the boundary, and for a boundary point $(a,0)$ choose a sufficiently small tangent disk, whose closure is tangent only at $(a,0)$. Thus $M$ is Hausdorff and hence $T_1$ by the two disjoint opens. The boundary $D=\mathbb{R}\times \{0\}$ is closed and discrete, while $E=\mathbb{Q}\times {\mathbb{Q}}_{>0}$ is countable and dense by the rational-density property.

Suppose the plane were normal. Then [L2] gives an injection $\mathcal{P}(D)\to \mathcal{P}(E)$. For the cardinal bridge in [L3], binary sequences map bijectively to the Cantor set and hence inject into $\mathbb{R}$; $x\mapsto \{q\in \mathbb{Q}:\iota (q)<x\}$ injects $\mathbb{R}$ into $\mathcal{P}(\mathbb{Q})$ by rational density; $\mathbb{Q}\approx \mathbb{N}$ transports the latter to $\mathcal{P}(\mathbb{N})$; and characteristic functions identify $\mathcal{P}(\mathbb{N})$ with binary sequences. Schröder--Bernstein therefore gives $\mathbb{R}\approx \mathcal{P}(\mathbb{N})$, and the projection transports this to $D\approx \mathcal{P}(\mathbb{N})$. By [L4], an injection $E\to \mathbb{N}$ induces an injection $\mathcal{P}(E)\to \mathcal{P}(\mathbb{N})$, while the just-established bijection gives $\mathcal{P}(D)\approx \mathcal{P}(\mathcal{P}(\mathbb{N}))$. Their composite is therefore an injection $\mathcal{P}(\mathcal{P}(\mathbb{N}))\to \mathcal{P}(\mathbb{N})$. The singleton map supplies the reverse injection, so [L5] makes this impossible.

The tangent-disk coordinate charts obtained by radial projection from the tangency point give metrizable neighbourhoods at boundary points; Euclidean disks do so above the boundary. To separate a point $p$ from a closed $F$ not containing it, first take a basic neighbourhood of $p$ disjoint from $F$. If $p=(a,0)$, choose a tangent disk $T_r(p)$ disjoint from $F$ and define $f(p)=1$ and, for $(u,v)$ with $v>0$, $f(u,v)=\max \text{\hspace{0.17em}⁣}\{0,1-\frac{(u-a{)}^2+v^2}{rv}\}\mathrm{.}$ Its support is the smaller tangent disk $T_{r\mathrm{/}2}(p)$, and $f\to 1$ at $p$ because $(u-a{)}^2+v^2<\epsilon rv$ is exactly membership in a sufficiently small tangent disk. It is ordinarily continuous above the boundary and zero on a tangent neighbourhood of every other boundary point, so it is continuous on $M$ and vanishes on $F$. If $p$ has positive height, an ordinary Euclidean bump supported in a small disk disjoint from $F$ and from the boundary has the same properties, extended by zero elsewhere. Thus $M$ is completely regular; with the $T_1$ conclusion of step 1.1, [L6] makes it Tychonoff and locally metrizable.

Hence the plane is not normal. If it were paracompact, its Tychonoff property gives Hausdorffness and [L5] would make it normal; if it were metrizable, Stone's theorem, under choice: every metric space is paracompact would make it paracompact. Both are impossible.

This proves the stated profile.