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$\newcommand\CT{\mathscr T}\newcommand\bbR{\mathbb R}\newcommand\bbI{\mathbb I}\newcommand\bbbI{\kern.4mm\mathbb I\kern.4mm}\newcommand\kp[1]{\kern.#1mm}$Specifically, is $\CT$ a first countable topology when $\CT$ is constructed in the following manner? Let $S$ be the set of all pairs $(f,g)$ where $f$ is a continuous function $\bbI=[\kp7 0\kp5,1\kp5]\to\bbR$ and $g$ is a continuous function defined precisely on the relatively open set ${\rm N}\kp5 f=\{\,t:f\kp5 t<0\,\}$ in $\bbI\kp5$, and such that it be a restriction of some function $h$ in $C\kp5(\bbbI)\,$. Let $q$ be the (quotient) map defined on the standard (uniform convergence) topological space $X=C\kp5(\bbbI)\times C\kp5(\bbbI)$ by $(\kp3 f\kp2,h\kp3)\mapsto(\kp3 f\kp2,h\,|\,{\rm N}\kp5 f\kp3)$ so that $q$ will be surjective $X\to S\kp4$. Then let $\CT$ be the strongest topology for $S$ such that $q$ be continuous $X\to(\kp2 S\kp3,\CT\,)\kp9$.

On the background: I am considering a stochastic well-posedness Navier−Stokes−Fourier model where the initial-boundary data set will have a structure partially analogous to that of $S$ above. I think I have a proof that $\CT$ be a Hausdorff topology, but then arose the question whether it be Polish or at least metrizable or first countable. Probably I would have found the answer myself after thinking about the matter long enough but I wanted to test whether the MO topology/functional analysis/real analysis community can find the answer more quickly.

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The space $S$ is not first-countable.

Let $f$ be given by $f(x)=0$ if $x\le\frac12$ and $f(x)=\frac12-x$ if $x\ge\frac12$, and let $h$ be the zero function. Then a pair $\langle f,g\rangle$ is equivalent to $\langle f,h\rangle$ iff $g(x)=0$ for $x\ge\frac12$. Define $g_n$ by $$ g_n(x)=\begin{cases} n & x\le\frac12-\frac1n\\ n-n^2(x-\frac12+\frac1n) & \frac12-\frac1n\le x\le\frac12 \\ 0 & \frac12\le x \end{cases} $$ Let $A$ denote the equivalence class of $\langle f,h\rangle$. Then $\langle f,g_n\rangle\in A$ for all $n$, but $\langle f-\varepsilon,g_n+\varepsilon\rangle\notin A$ for all $\varepsilon>0$.

There is no countable local base at the point $A$ of $S$. Let $\langle U_n:n\in\mathbb{N}\rangle$ be a sequence of neigbourhoods of the point $A$ in $S$. Look at the preimages $q^{-1}[U_n]$, where $q$ is the quotient map. For each $n$ there is an $\varepsilon_n<\frac13$ such that the pair $p_n=\langle f-\varepsilon_n, g_n+\varepsilon_n\rangle$ is in $q^{-1}[U_n]$. Note that, because $f-\varepsilon_n$ is negative everywhere the pair $p_n$ is equivalent only to itself. It follows that $P=\{p_n:n\in\mathbb{N}\}$ satisfies $P=q^{-1}[q[P]]$.

Also: $P$ is closed: the pairs $\langle f,g_n\rangle$ are at least distance $1$ apart, hence the pairs $p_n$ are at least $\frac13$ apart, so $P$ is closed and discrete. We find that $q[P]$ is a closed set in $S$, and $A\notin q[P]$. But $q(p_n)\in q[P]\cap U_n$ for all $n$.

Thus the $U_n$ do not form a local base at $A$.

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  • $\begingroup$ I made some edits since some points seemed weird. If it is now what you actually meant to be a proof that $\mathscr T$ is not first countable, I will accept it. $\endgroup$ Commented Oct 20 at 12:27
  • $\begingroup$ @TaQ I made some further edits. The first sentence says what the answer actually is. And I did make some weird errors indeed. $\endgroup$ Commented 2 days ago

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