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I want to show that $(\mathbb{R}, \mathcal{T}_{\text{cocoun}})$ is not second- countable as we know that all space second-countable is a first-countable it is enough to see that space is not first-countable.

$\mathcal{T}_{\text{cocoun}}$ is a co-countable topology.

By contradiction, suppose that $(\mathbb{R}, \mathcal{T}_{\text{cocoun}})$ is first-countable, let $p\in \mathbb{R}$and $\mathcal{B}(p)=\{B_{1}, B_{2}, \dots, B_{n}, \dots \}$ a local basis countable for $p$. note that $$B_{1}^{c}, B_{2}^{c}, B_{3}^{c}, \dots, B_{n}^{c}, \dots \hspace{0.3cm} \text{are countable set}$$ And $$\bigcup_{i\in \mathbb{Z^{+}}}B_{i}^{c}\subseteq \mathbb{R} \hspace{0.3cm} \text{is a countable union}$$

As $\mathbb{R}$ is not contable, there exists $\alpha \in \mathbb{R}$ s.t. $$\alpha \notin \bigcup_{i\in \mathbb{Z^{+}}}B_{i}^{c} \hspace{0.3cm} \text{y} \hspace{0.3cm} \alpha \neq p$$ For this, we have that $$\alpha \in \bigcap_{i\in \mathbb{Z^+}}B_{i} \hspace{0.3cm} \implies \hspace{0.3cm} \alpha \in B_{i} \hspace{0.3cm} \text{for all} \hspace{0.3cm} i\in \mathbb{Z^+}$$

So $\mathbb{R}-{\alpha} \in N(p)$, and exists $k \in \mathbb{Z^+}$ s.t. $$B_{k}\subseteq \mathbb{R}-{\alpha}$$

As $\alpha \in B_{k}$ then $\alpha \in \mathbb{R}-\{\alpha \}$, but this is a contradiction.

Maybe this could work, by making some adjustments? any suggestions? I appreciate it!

Wrloord
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1 Answers1

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Your attempt and idea is very good and very nearly complete. Instead of just thinking about a particular $\alpha$, though, we want to exhibit an open set which does not contain $B_j$ for any $j$. This would contradict (second) (first) countability; notice that your observations never really require that $p\in B_j$ for all $j$, this proof idea works with or without a neighbourhood basis. However, to prove it is not first countable is a stronger statement, so let's do that.

Take $(B_j)_{j\in\Bbb N}$ a supposed local basis at $p$. $K:=\bigcup_{j\in\Bbb N}(\Bbb R\setminus B_j)$ is countable and does not contain $p$; $\Bbb R\setminus K$ is cocountable and does contain $p$. Oh, so $\Bbb R\setminus K$ is actually an open neighbourhood of $p$! Yes it was true there are $\alpha\in\Bbb R\setminus K$ but the important thing is that the whole of $\Bbb R\setminus K$ is open. Clearly $\Bbb R\setminus K$ does not contain $B_j$ for any $j\in\Bbb N$, so $(B_j)_{j\in\Bbb N}$ cannot be a local basis.

FShrike
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