Show that $Y$ is not first-countable at $[0]$

Question

(a) Show that $X := \{0, 1\}^{\mathbb R}$ is not first-countable.

(b) Let $X := \bigcup_{n=0}^∞ [2n, 2n + 1]$ be a union of countably many disjoint closed intervals. Let $Y$ be the quotient of $X$ obtained by gluing all the points $0, 2, 4, 6, \ldots$ together.

Show that $Y$ is not first-countable at $[0]$ (the equivalence class of $0$). Use a diagonalization argument.

My attempt:

$(a)$ I proceeded indirectly by showing that there exists a real-valued function on some subspace of $[0,1]^{\mathbb R}$ that is sequentially continuous but not continuous. It is the case of $F(f) = \int_0^1 f(t) \, dt$ where $f$ is measurable.

$(b)$ But I don't know how to prove this part. Any help please?

See this answer, which gives a fairly significant hint for proving that a certain subspace of your $X$ is not first countable at $[0]$. The same argument works here. — Brian M. Scott, Nov 20 '20 at 23:06
Twice the same question: https://math.stackexchange.com/q/3915809 — Paul Frost, Nov 21 '20 at 12:11

score 1 · Answer 1 · answered Nov 21 '20 at 05:28

For (b): If $V=\{U_j:j\in \Bbb N_0\}$ is a countable set of nbhds of $[0]$ in $Y$ then for each $j$ there is a function $f_j:\Bbb N_0\to (0,1)$ such that $ U_j\supset (2n,2n+f_j(n))$ for every $n.$ Then $[0]\cup(\cup_{n\in \Bbb N_0}(2n, f_n(n)/2)\,)\,)$ is a nbhd of $[0]$ which is not a subset of any member of $V.$

Show that $Y$ is not first-countable at $[0]$

1 Answers1