An example of a $T_{3}$ topological space that is not $T_{4}$

Question

I have found this example on the internet here http://mathonline.wikidot.com/a-t3-space-that-is-not-a-t4-space , but I need a simpler one, is there any suggestions?

Answer 1

You can query $\pi$-base for such questions. It lists more than 10 such spaces. Your link is among them as Sorgenfrey's Half-Open Square Topology (I usually call it the Sorgenfrey plane). The rational sequence topology is one of my favourites in terms of simplicity. Mrowka's $\Psi$-space is also very nice, but requires a bit more set theory, as does the Tychonoff-plank.

A simple variation on the Tychonoff plank also works:

let $X = \mathbb{N} \cup \{\infty_1\}$, be the one-point compactification of the natural numbers (so points of $\mathbb{N}$ are still isolated and neighbourhoods of $\infty_1$ are of the form $\{\infty_1\} \cup \mathbb{N}\setminus F$, where $F \subset \mathbb{N}$ is finite. And let $D$ be the reals (or any other uncountable set) in the discrete topology, and let $Y = D \cup \{\infty_2\}$ by its one-point compactification (so neighbourhoods of $\infty_2$ have the same form as those of $\infty_1$, replacing $\mathbb{N}$ with $D$, of course. Then give $X \times Y$ the product topology, and let our space be $Z = X \times Y\setminus \{(\infty_1, \infty_2)\}$. Then $A = \mathbb{N} \times \{\infty_2\} \subseteq Z$ is closed (as the removed point is its only limit point in $X \times Y$) and similarly, so is $B = \{\infty_1\} \times D \subseteq Z$ and these sets are disjoint. $Z$ is $T_3$ as both $X$ and $Y$ are normal (being compact and Hausdorff) so $T_3$ as well and $T_3$ is preserved by products and subspaces.

Suppose there are open sets $U,V \subseteq Z$ such that $A \subseteq U$ and $B \subseteq V$. For each $n \in \mathbb{N}$, $(n, \infty_2) \in A \subseteq U$, so we can find $F_n \subseteq D$ such that

$$\{n\} \times (\{\infty_2\} \cup (D \setminus F_n)) \subseteq U\text{,}$$ using the local bases and the fact we are in a product. Now $\cup_n F_n$ is at most countable in $D$, so there is some $d \in D$ which is not in any $F_n$. Here we use that $D$ is uncountable. Then $(\infty_1, d) \in V$ so we find a finite subset $F \subseteq \mathbb{N}$ such that

$$(\{\infty_1\} \cup (\mathbb{N} \setminus F)) \times \{d\} \subseteq V\text{.}$$

Then pick $n_0 \in \mathbb{N}\setminus F$, which is clearly possible. But then $(n_0, d) \in U \cap V$ as $n_0 \in \mathbb{N}\setminus F$, so $(n_0, d) \in (\{\infty_1\} \cup (\mathbb{N} \setminus F)) \times \{d\} \subseteq V$ and $d \in D\setminus F_{n_0}$ so $(n_0, d) \in \{n_0\} \times (\{\infty_2\} \cup (D \setminus F_{n_0})) \subseteq U$. So $U \cap V \neq \emptyset$ and so $A$ and $B$ cannot be separated in $Z$.

Answer 2

Niemytzki's Tangent Disc Topology

This is a topology that is regular (i.e. $T_3$) but not normal (i.e. not $T_4$).

Say $P := \{ (x,y)\in\mathbb{R}^2 | y > 0 \}$ is the open upper half plane with Euclidean topology $\tau$.

Say $L:=\mathbb{R}$ is the real axis.

Say $X:=P \cup L$. Define a topology $\tau^*$ for $X$ by adding to $\tau$ all sets $\{ x \} \cup D$ where $x \in L$ and $D$ is an open disc in $P$ that is tangent to $x$.

Then you can show that $(X,\tau^*)$ is regular, see e.g. Steen & Seebach, Counterexamples in topology, example 82, 2. (availlable via google as pdf).

However it is not normal: (1) Every subset of $L$ must be closed, since every point $x$ in $L$ is the complement of $\bigcup_{y\neq x}\bigcup_{\epsilon > 0}D_\epsilon^y$ of a union of $D_\epsilon^y:=D_\epsilon \cup \{ y \}$. (2) Thus the rationals $\mathbb{Q}$ and irrationals $\mathbb{I}$ are two closed subsets in $X$ and these can not be separated in $X$.

Answer 3

(I). For $n\leq 3\frac {1}{2}$ the Tychonoff product of a family of $T_n$ spaces is $T_n,$ and a subspace of a $T_n$ space is a $T_n$ space. A linear space is a $T_3$ space. (These are easy to prove).

(II). So with the $\in$-order topologies on $\omega+1$ and on $\omega_1+1,$ (that is $x<y\iff x\in y$), the space $X=[(w+1)\times (w_1+1)]$ \ $\{(w,w_1)\}$ is a $T_3$ space.

$X$ is not a normal space: Let $A=\omega\times \{\omega_1\}$ and $B=\{\omega\}\times \omega_1.$ Observe $A$ and $B$ are closed in $X$ and $A\cap B=\phi.$

Now let $U$ be open in $X$ with $U\supset A.$ We show that $B\cap Cl_X(U)\ne \phi.$

For each $n\in \omega$ let $g(n)$ be the least (or any) $x\in \omega_1$ such that $U\supset \{n\}\times \{y\leq \omega_1: y>x\}.$ Let $G=(\sup_{n\in \omega}g(n))+1.$ Then $G<\omega_1.$

For brevity let $G^*=\omega \times \{y\in \omega_1: y\geq G\}.$ We have $G^*\subset U. $

But $(\omega, G)\in B\cap Cl_X(G^*)\subset B\cap Cl_X(U).$ Because for any open $V$ in $X$ with $(\omega, G)\in V,$ there exists $n\in \omega$ with $$V\supset \{x\in \omega+1:x>n\}\times \{G\}\supset \{(n+1,G)\}\in G^*\subset U. $$

So $B\cap Cl_X(U)\ne \phi$ for any open $U$ of $X$ such that $A\supset U$. So $A$ and $B$ cannot be completely separated in $X.$