I read somewhere the following stament:

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

Can someone give an example or a reference?

  • 2
    $\begingroup$ In the hope of avoiding an explosion of variants of Stefan's proof, I note that his and Arthur's proofs are both special cases of the following (and there are lots more special cases). Take a countable set $S$ and a filter $F$ on S having no countable base and containing all cofinite subsets of $S$. Topologize $S\cup\{p\}$ (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
  • $\begingroup$ The Appert space, pointed out in Ramiro de la Vega's answer, also fits the description in my previous comment. $\endgroup$ – Andreas Blass Aug 22 '12 at 23:34

A search in Spacebook for Normal + $T_1$ + Countable + not Second-Countable gives three spaces:

"Single Ultrafilter Topology" which is the one described in Stefan´s answer,

"Arens-Fort Space" given in Arthur´s answer and

"Appert Space" which you can find here.


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.


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 $i$.


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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.