# A continuous image of a second countable space which is not second countable.

Construct an example proving that a continuous image of a second countable space may be not second countable.

I construct an example by taking two different topology on $I=[0,1]$, $(I,\mathcal{X})$ and $(I,\mathcal{Y})$ where $\mathcal{X}$ is the standard one, and $\mathcal{Y}$ is generated from the base of the open interval with end points in Cantor set. The map is the identity map.

Is my construction right? Is there any other constructions?

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Perhaps you could elaborate on why $(I, \mathcal{Y})$ is not second countable. – Nate Eldredge Apr 24 '12 at 13:37

Take $f: (\mathbb R, T_{Euclid}) \to (\mathbb R, T_{cof})$ the identity map from the standard topology to the cofinite topology on $\mathbb R$.

To see why a basis for the cofinite topology on $\mathbb R$ cannot be countable (by contradiction) assume that you have a countable basis $B$ of $T_{cof}$. Then for every $O$ in $B$, $O^c = \mathbb R \setminus O$ is finite (by definition). Take the union $\bigcup_{O \in B} O^c$ over all $O^c$ in $B$. Then $\bigcup O^c$ is countable hence a proper subset of $\mathbb R$. Pick a point $x$ in $\mathbb R \setminus \bigcup O^c$. Then $\mathbb R \setminus \{x\}$ is open in the cofinite topology but you cannot write it as the union of sets in $B$. Hence $B$ cannot be a basis.

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I didn't noticed that the cofinite topology is not second countable. Is it possible to construct a example maps into the dictionary topology of $\mathbb{R}^2$? – Minghao Liu Apr 24 '12 at 13:54
@pirriperdos I don't know what the dictionary topology is. To see why the cofinite topology cannot have a countable basis see the added part above. – Rudy the Reindeer Apr 24 '12 at 14:24
Dictionary topology for $\mathbb{R}^2$ is the topology generated from the dictionary order of $\mathbb{R}^2$, $(a_1,b_1)<(a_2,b_2)\leftrightarrow a_1<a_2\lor (a_1=a_2\land b_1<b_2)$. – Minghao Liu Apr 24 '12 at 14:57
@pirriperdos I had to look it up: Apparently, the dictionary topology on $\mathbb R \times \mathbb R$ is the same as $\mathbb R_{discrete} \times \mathbb R$. So this is finer than $\mathbb R \times \mathbb R$ and it won't have a countable basis. Unfortunately, you can't take the identity map in this case since it wouldn't be continuous. As for other constructions: I don't know off the top of my head. I hope someone else can answer this. – Rudy the Reindeer Apr 24 '12 at 15:14

This is exercise 16B.1 in Willard's General Topology: For each $n\in \Bbb N$, let $I_n$ be a copy of $[0,1]$. Let $X$ be the disjoint union of the spaces $I_n$. Identify the left hand endpoints of all the $I_n$ and let $Z$ be the resulting quotient space. The distinguished point in $Z$ has no countable nhood base, though $X$ is second countable.

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I thought this was an excellent example. It seems really important that countable products of countable sets are uncountable (except, of course, when there are only finite $n$'s such that $|A_n| \neq 1$). – Doug May 8 '14 at 14:28

How about a separable Banach space $X$, with $\tau$ the norm topology and $\tau_w$ the weak topology. The identity map $(X, \tau) \to (X, \tau_w)$ is continuous, and $(X,\tau)$ is a separable metric space, hence second countable, but $(X, \tau_w)$ is not even first countable.

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I cannot understand your construction now because lack of knowledge, sorry for that. – Minghao Liu Apr 24 '12 at 13:52
In a similar vein but maybe more elementary: take $C[0,1]$ with the sup-norm and the topology of pointwise convergence. – t.b. Apr 24 '12 at 13:53
@t.b. How do we know that $C([0,1])$ is not second countable with the topology of pointwise convergence? It's clear to me that it isn't if we take all functions $[0,1] \to \mathbb R$ since then we get the product topology but by restricting to continuous functions only we are removing quite a few points from the space so we're left with a smaller space that could have a smaller base, no? – Rudy the Reindeer Apr 25 '12 at 15:32
@MattN. If it were 1st countable, it'd be metrizable. Let $d$ be a metric on $C = C[0,1]$ s.t. $f_n \to 0$ ptw. iff $d(f_n,0) \to 0$. A) Show: For all $x \in [0,1]$ we have $$\delta(x,n) = \sup{\{|f(x)|\,:\,f \in C,\,d(f,0) \lt 1/n\}}\xrightarrow{n\to\infty} 0.$$ B) Let $A_n = \{x \in [0,1]\,:\,\delta(x,n) \lt 1/2\}$. Then $[0,1] = \bigcup_n A_n$ by A), so some $A_{N}$ contains infinitely many points $\{x_k\}_{k \in \mathbb{N}}$. C) Choose p.w. disjoint open $U_k \ni x_k$ and $f_k$ with $|f_k(x_k)| > 1/2$ and support in $U_k$. Then $d(f_k,0) \geq 1/N$ for all $k$, while $f_k \to 0$ ptw... – t.b. Apr 25 '12 at 17:03
@t.b. Not too terse (for now). Thank you! – Rudy the Reindeer Apr 25 '12 at 17:26