The answer https://math.stackexchange.com/a/698429/32337 purports to prove that Mysior's regular topological space $X$ is not completely regular. But the crucial part of the proof, consisting of an induction allegedly showing that the sets $B_n$ defined there are infinite, seems to be riddled with difficulties in both notation and substance and is somewhat tangled in the order of steps. (See my comments there.)
Can somebody supply a complete and correct proof of that induction?
Here again is the example and the proof up to the inductive step at issue. (Mysior's original example used $\mathbb{R} \times [0, \infty)$, but as in https://math.stackexchange.com/a/698429/32337 nothing really changes by using $\mathbb{R} \times [0, 2)$, instead.)
Let $H = \mathbb{R} \times (0, 2)$ and $L = \mathbb{R} \times \{0\}$. Define $X = H \cup L \cup \{a\}$ to be the set obtained by adjoining a new point $a$ to $H \cup L$ and provide $X$ with the topology for which:
- each $(x, y) \in H$ is isolated;
- at each $(x, 0) \in L$, a local base consists of all subsets of $\{(x,0)\} \cup I_{x} \cup J_{x}$ that include almost all, but not all, points from $I_{x} \cup J_{x}$, where $I_{x}$ and $J_{x}$ are the vertical and slanted "whiskers" given by $I_{x} = \{(x, y) : 0 \leq y < 2\}$ and $J_{x} = \{(x + y, y) : 0 \leq y < 2\}$, respectively; and
- at $a$ a local base consists of the sets $U_{n} = \{a\} \cup \{(x, y): n < x, 0 \leq y< 2\}$ for $n = 1, 2, 3, \dots$.
The resulting space $X$ is regular because each basic neighborhood of each point of $H \cup L$ is closed in $X$ and $\mathop{\textrm{cls}} U_{n+2} \subset U_{n}$ for each $n$.
Just suppose that $X$ were completely regular. Then there must exist a continuous $f : X \to [0, 1]$ with $f(a) = 0$ and $f(z) = 1$ for all $z \in E$, where $E$ is the closed set given by $E = \{(x, 0) : 0 \leq x \leq 1\}$. Since $f(a) < 1$, there is some $m$ for which $f(U_{m}) \subset [0, 1)$.
For $n = 1, 2, 3, \dots$, define $B_{n} = \{x : n-1 < x \leq n , f(x, 0) = 1\}$. We are going to show that $B_{n}$ is infinite for every $n$. In particular, $B_{m+1}$ will be infinite, so there will be some $x \in B_{m+1}$. This will yield a contradiction because, on the one hand, $(x, 0) \in U_{m}$ so that $f(x, 0) = 1$ by definition of $B_{m+1}$, and, on the other hand, $f(x, y) < 1$ since $(x, 0) \in U_{m}$.
Question 1: Is that last sentence correct — all that is actually used about the sets $B_n$ to get the contradiction is that they are nonempty, and their infinitude is not actually used?
We use induction to prove that $B_{n}$ is infinite for each $n$. First, $B_{1}$ is infinite because $(0, 1] \times \{0\} \subset E$ implies that $(0, 1] \subset B_{1}$.
Overall question: How does the inductive step get completed? Here's how far I can get:
Now let $n \geq 1$ and assume that $B_{n}$ is infinite. Then $B_{n}$ has a denumerable subset $D$. Let $d \in D$. Now $J_d$ is closed in $X$, and by continuity of $f$ the inverse image $f^{-1}(1)$ is a $G_{\delta}$-set. Hence $J_d \setminus f^{-1}(1)$ is an $F_{\sigma}$-set not containing $(d, 0)$.
Question 2: Why is $J_d \setminus f^{-1}(1)$ countable — what guarantees that at least one of the sets in the union is nonempty?
Note A: Mysior's proof actually works with an $f$ for which $f(z) = 0$ on $E$ and then of course uses $f^{-1}(0)$ where I use $f^{-1}(0)$ above. But nothing essential changes from the above.
Note B: Instead of a completion of Mysior's proof whose skeleton is given above, I would be happy to see instead a complete and correct completion of the proof given in https://math.stackexchange.com/a/698429/32337.
