I read somewhere the following stament:

There are countable normal $T_1$ spaces of uncountable weight.

Can someone give an example or a reference?

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I read somewhere the following stament:

There are countable normal $T_1$ spaces of uncountable weight.

Can someone give an example or a reference?

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Let $p$ be an ultrafilter over $\mathbb N$ that contains all cofinite sets. Let $X=\mathbb N\cup\{p\}$. For each $n\in\mathbb N$ let $\{n\}$ be an open set. For each $A\in p$ let $A\cup\{p\}$ be an open set. Consider the topology generated by those open sets. Notice that a subset of $X$ is open iff it is a subset of $\mathbb N$ or contains the point $p$ and includes a set $A\in p$.

Since the ultrafilter is not generated by a countable set, the point $p$ has no countable neighborhood basis. It follows that the space is of uncountable weight.

Now let $A,B\subseteq X$ be disjoint and closed.

By the classification of the open subsets of $X$, every set containing $p$ is closed.
A set that does not contain $p$ is closed iff it is disjoint from some set in $p$.

Since $A$ and $B$ are disjoint, at most one of them contains $p$. Wlog let $B$ be the set that does not contain $p$. Then $B$ is open. Since $B$ is closed, $X\setminus B$ is open. Since $A\subseteq X\setminus B$, $B$ and $X\setminus B$ are open sets separating $B$ and $A$. It follows that $X$ is normal.

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Another classical example is the Arens-Fort Space.

Let $X = \omega \times \omega$. For each $A \subseteq X$ and $n \in \omega$ we let $A_n = \{ m \in \omega : (n,m) \in A \}$ denote the $n$th section of $A$.

Topologise $X$ by taking each point of $X \setminus \{ (0,0) \}$ to be isolated, and let $U \subseteq X$ be a neighbourhood of $(0,0)$ iff it contains $(0,0)$ and all but finitely many sections of $U$ are co-finite.

Clearly this space is T$_1$. If $F, E \subseteq X$ are disjoint closed sets, then one, say $E$, does not contain $(0,0)$. So $E$ is clopen, and $X \setminus E$ is an open set including $F$ which is disjoint from $E$. Thus $X$ is normal.

To show that $X$ is not second countable, it suffices to show that there is no countable base at $(0,0)$. If $\{ U^{(i)} \}_{i \in \omega}$ is any family of open neighbourhoods of $(0,0)$, we inductively define a sequence of pairs of natural numbers $\{ (n_i,m_i) \}_{i \in \omega}$ so that:

- $n_i > n_{i-1}$ is such that $U^{(i)}_{n_i}$ is non-empty (say $n_{-1} = 0$); and
- $m_i \in U^{(i)}_{n_i}$.

Then $V = X \setminus \{ ( n_i , m_i ) : i \in \omega \}$ is an open neighbourhood of $(0,0)$, and by construction $U^{(i)} \not\subseteq V$ for all $i$.

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Another favourite example of mine: the Hewitt-Marczewski-Pondiczery theorem says that the product of continuum or fewer separable spaces is separable. So $\mathbb{R}^\mathbb{R}$ has a countable dense subset $D$ and it is clear any countable regular space is normal (from being Lindelöf), and it's not hard to show that every point of $D$ has a local base of $\mathfrak{c}$ many open sets (and not fewer). So it's not like the other examples mentioned, in that there are no isolated points.

`$S\cup\{p\}$`

(where $p\notin S$) by making a subset $U$ open if either $p\notin U$ or $(U\cap S)\in F$. Then $X$ is normal because of any two disjoint (closed) sets one is open. There is no countable base for the topology because there is no countable neighborhood base at $p$. $\endgroup$ – Andreas Blass Aug 21 '12 at 15:34