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.
 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 $\mathbb{R}^2$? – pirriperdos 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. – Matt N. Apr 24 '12 at 14:24 Dictionary topology for $\mathbb{R}^2$ is the topology generated from the dictionary order of $\mathbb{R}^2$, $(a_1,b_1)<(a_2,b_2)\leftrightarrow a_1 How about a separable Banach space$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)$is not even first countable. - I cannot understand your construction now because lack of knowledge, sorry for that. – pirriperdos Apr 24 '12 at 13:52 In a similar vein but maybe more elementary: take$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$since then we get the product topology but by restricting to continuous functions only we are removing quite a few points from the space so we're left with a smaller space that could have a smaller base, no? – Matt N. Apr 25 '12 at 15:32 @MattN. If it were 1st countable, it'd be metrizable. Let$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! – Matt N. Apr 25 '12 at 17:26 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.