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The Wikipedia definition of a First-countable space states the following.

For each point $x$ in $X$ there exists a sequence $N1,N2,\dots$ of neighbourhoods of $x$ such that for any neighbourhood $N$ of $x$ there exists an integer $i$ with $N_i$ contained in $N$.

This seems like it will always be satisfied? Let $x\in X$ and let $N$ be a neighbourhood of $x$. Well then just put this $N$ into the sequence $N_1,N_2,\dots$, and of course $N \subset N$.

In effect, we can just make the sequence contain all neighbourhoods of $x$ and then of course there will exist

a sequence $N1,N2,\dots$ of neighbourhoods of $x$ such that for any neighbourhood $N$ of $x$ there exists an integer $i$ with $N_i$ contained in $N$.

So what am I missing about this definition?

José Carlos Santos
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ManUtdBloke
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    The set of neighborhoods of $x$ is not necessarily countable. – platty Aug 03 '17 at 15:50
  • Try adding uncountably many such $N$'s to your sequence... – b00n heT Aug 03 '17 at 15:52
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    Consider an uncountable set with the cofinite topology for a concrete example. – Daniel Fischer Aug 03 '17 at 15:58
  • Possible duplicate of Example of a topological space which is not first-countable – Danu Aug 03 '17 at 16:01
  • @Danu Not really, since there are first countable spaces where every point has uncountably many open neighbourhoods. The real line, for example. Sure, the existence of non-first-countable spaces is important here, but even in the context of first-countable spaces the OP's suggestion is flawed. – user642796 Aug 03 '17 at 16:16
  • Fair enough, @arjafi. The linked post answers the question in the title, at least. – Danu Aug 03 '17 at 16:24
  • @b00nheT But why should I do that? Why can't I just put a countable number of $N$'s in the sequence, one of which is equal to the original neighborhood? The definition just states that it requires the existence of a sequence, so why I can I not just take a countable sequence? – ManUtdBloke Aug 03 '17 at 17:12
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    @eurocoder Because the sequence is not supposed to depend on the given neighborhood $N.$ The sequence is supposed to depend only on the point $x.$ In English, that's clear from the syntax: "For each point $x$ in $X$ there exists a sequence $N_1,N_2,\dots$" – bof Aug 03 '17 at 17:50
  • To complement @bof's comment above, what the definition says can be written symbolically as, $$\forall x\color{red}{\exists\mathcal{B}}\color{blue}{\forall N}\exists N_i(N_i\in \mathcal{B}\land N_i\subseteq N)$$On the other hand what you are saying is, $$\forall x\color{blue}{\forall N}\color{red}{\exists\mathcal{B}}\exists N_i(N_i\in \mathcal{B}\land N_i\subseteq N)$$where $\mathcal{B}$ denotes a countable set of neighbourhoods of $x$. –  Aug 04 '17 at 04:08

2 Answers2

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No, you cannot make a sequence with all the neighbourhoods of a point $x$, e.g. there are as many neighbourhoods of $0$ in the real line as there are real numbers, for every $r > 0$, we have $O_r = (-r,r)$ which is a neighbourhood of $0$. And Cantor's diagonal argument shows that we cannot put the real numbers in a sequence.

The whole point of being first-countable is that there is for every $x \in X$ a fixed (you cannot change it, and add your $N$!), pre-given, (countable!) sequence of neighbourhoods $N_1, N_2, N_3, \ldots$ of $x$ such that every arbitrary neighbourhood of $x$ contains one of them. This allows us to essentially understand all neighbourhoods of a point from only countably many of them. This allows you to work with sequences and sequential continuity.

All metric spaces are first-countable because we can use $B(x, \frac{1}{n})_{n \in \mathbb{N}}$ as a countable local base for $x$. So the standard examples are first countable, but there are many non-first-countable spaces.

Henno Brandsma
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A space is first-countable if for each point $x$ there is a single sequence of neighborhoods such that every neighborhood of $x$ contains in some neighborhood in the sequence. Although for each specific neighborhood $V$ we can easily find a sequence with some element contained in $V$, we can't necessarily find a single sequence which works for every such neighborhood. (This is similar to the reason why the reals are uncountable, even though any specific real can be put on a list.)

This just shows why first-countability is not obviously trivial. It turns out that it is really nontrivial: there are lots of non-first-countable sequences. For example, consider the order topology on $\omega_1+1$. Every open neighborhood of $\omega_1$ here contains a neighborhood of the form $(\alpha, \omega_1]$ for $\alpha$ a countable ordinal. But the supremum of countably many countable ordinals is countable, so for any sequence $U_i$ ($i\in\mathbb{N}$) of neighborhoods of $\omega_1$ contains a neighborhood of the form $(\beta, \omega_1]$ for some countable ordinal $\beta$. But then the neighborhood $(\beta+1, \omega_1]$ doesn't contain any of the neighborhoods $U_i$.

EDIT: here's a much better example, not requiring ordinals at all: the cocountable topology on any uncountable set $S$. For any sequence of neighborhoods $(U_i)_{i\in\mathbb{N}}$ of a point $x$, the intersection $\bigcap U_i$ is again cocountable, and hence uncountable (since $S$ is uncountable). Pick $s\in \bigcap U_i$, $s\not=x$; then $(\bigcap U_i)\setminus\{s\}$ is a neighborhood of $x$ not containing any of the $U_i$s.

Interestingly, first countability is nontrivially nontrivial: showing that a space is not first countable generally requires using a principle of the form "the union of countably many "small" sets is "small,"" and such principles are generally not true without some amount of the axiom of choice. For instance, it is consistent with ZF that every infinite set is a countable union of sets of strictly smaller cardinality! In fact, I don't offhand know of an example of a space which is provably non-first-countable in ZF alone (and I've now asked a question about this).

(Incidentally, there are some neat examples if choice fails sufficiently badly - e.g. the cofinite topology on any amorphous set, if such a set exists.)

Noah Schweber
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    I think that the first paragraph of your answer may be made more concise by simply pointing out that OP misinterprets the order of quantification of variables (as I commented below the post itself). –  Aug 04 '17 at 04:20
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    @user170039 While more concise, I think that natural language may clarify things for the OP faster. – Noah Schweber Aug 04 '17 at 04:22