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We define the following topology on the reals; $T = \{ U \subset\mathbb{R}\:|\:0\notin U\text{ or }\mathbb{R}\setminus U\text{ finite}\} $. Is this topology first countable? I'm trying to find a local basis at each point $x \in \mathbb{R}$ but I'm confused with the topology having cofinite open sets. What would be a correct approach?

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up vote 4 down vote accepted

For $x \neq 0$, $\{\{x\}\}$ is a fundamental system of neighborhoods of $x$. Suppose that $0$ admits a countable fundamental system of neighborhoods: $U_n= \mathbb{R} \backslash F_n$, where $F_n$ is finite. Because, $\bigcup\limits_{n \geq 0} F_n$ is countable, there exists $p \notin \bigcup\limits_{n \geq 0}F_n$ with $p \neq 0$ ; here, $\mathbb{R} \backslash \{p\}$ is an open neighborhood of $0$ but $U_n \nsubseteq \mathbb{R} \backslash \{p\}$ for all $n \geq 0$.

So your space is not first countable, but $0$ is the only problematic point.

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$\mathbb{R} \backslash \{p\} \nsubseteq U_n$ near the end should be the other way around. – Ayman Hourieh Jan 13 '13 at 14:39
@AymanHourieh: Thank you, I corrected. – Seirios Jan 13 '13 at 14:41
@Seirios: Exactly! Because without the cofinite sets it would be the excluded-point topology which is first countable, right? – omar Jan 13 '13 at 14:43
@omar: Yes, without the cofinite sets, the only open neighborhood of $0$ would be the whole space. – Seirios Jan 13 '13 at 14:46

No. it is not first-countable.

Assume $x$ has countable local basis $\{B_1, B_2,\cdots\}$, $B_i = \mathbb{R}\setminus \{ x_{i1},x_{i2},\cdots , x_{in_i}\}$. We can consider $$A=\bigcup_{i=1}^\infty \mathbb{R}\setminus B_i \cup \{x\}.$$ Then $A$ is countable. So, there exists $y$ such that $y\notin A$. And $\mathbb{R}\setminus \{y\}$ is open neighborhood of $x$. But $B_i \not\subset \mathbb{R}\setminus \{y\}$ for all $i$.

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