Is there a Tychonov space $(X,\mathcal T)$ with $a,b\in X$ such that $a$ has a countable neighborhood basis while $b$ does not have any countable neighborhood bases?
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the answer seems to be positive. I think such space can be made with attaching two spaces. But I'm not sure if it is possible. – Minimus Heximus Sep 05 '14 at 09:45
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1Just take a topological sum of two Tychonoff spaces, such that only one of them is first countable. – Martin Sleziak Sep 05 '14 at 10:56
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@MartinSleziak By sum do you mean a product? Do you have an example of a Tychonov space which is not first countable? – Minimus Heximus Sep 05 '14 at 10:59
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1I'm pretty sure @MartinSleziak is talking about topological sum, also called the disjoint union, and, for the category-theory-literate, coproduct. – user642796 Sep 05 '14 at 11:14
2 Answers
Here are a couple more useful examples.
The "one-point compactification of an uncountable discrete space." Let $X$ be an uncountable set containing a chosen element $\infty$. The open subsets of $X$ are exactly those $U \subseteq X$ such that either $\infty \notin U$ or $X \setminus U$ is finite. This is a compact Hausdorff space, and so it's normal. Each point $x \neq \infty$ is isolated, and so $\{\,\{x\}\,\}$ is a countable (finite) neighborhood base at $x$. However there is no countable neighborhood base at $\infty$: if $U_1, U_2 , \ldots$ are open neigborhoods of $\infty$, then $\bigcup_i ( X \setminus U_i ) = X \setminus \bigcap_i U_i$ is countable, meaning that $\bigcap_i U_i$ is uncountable, and so if $x \in \bigcap_i U_i$ is different from $\infty$ we have that $V = X \setminus \{ x \}$ is an open neighborhood of $\infty$, but $V \not\subseteq U_i$ for all $i$.
The closed ordinal space $[0,\omega_1]$, where $\omega_1$ is the least uncountable ordinal. Again this is a compact Huasdorff space, so it is normal. In this space each point $\alpha \in [ 0 , \omega_1 ) = [0 , \omega_1 ] \setminus \{ \omega_1 \}$ has a countable neighborhood base, however the point $\omega_1$ does not. (These facts are largely due to set-theoretic properties of ordinal numbers.)
To get the desired counterexample, you can simply take a topological sum of two Tychonoff spaces such that only one of them is first countable.
An example of a Tychonoff space, which is not first countable, is the product $I^I$ of $\mathfrak c$-many copies of unit interval $I=[0,1]$. Since it is compact Hausdorff, it is also Tychonoff.
It is not first-countable, see here: Uncountable Cartesian product of closed interval
The same counterexample was used here: Topological counterexample: compact, Hausdorff, separable space which is not first-countable
To find other such examples you could search here or simply Google for some reasonable keywords. Or you could have a look at Counterexamples in Topology. It is great book for purposes of finding counterexamples and, if I am not mistaken, it served as a basis for the database I linked to.

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