T_D spaces

Definition

Again, we are going to do things differently from our main reference,

📚 Jorge Picado & Aleš Pultr Frames and Locales: topology without points. Springer Basel, Series Frontiers in Mathematics, Vol. 28, xx + 400 pp., 2012 (ISBN: 978-3-0348-0153-9).

There, the definition is slightly complicated. There is also section 1.4 dealing with some even more complicated characterizations of $T_D$ spaces. I think it would be nice if this very simple characterization given here was present in chapter 1.

A topological space $(X, \tau)$ is a $T_D$ space when for every $x \in X$ , $x$ is an isolated point in $\overline{\{x\}}$ .

Another way to say that is:

A topological space $(X, \tau)$ is a $T_D$ space when for every $x \in X$ , $\overline{\{x\}} \setminus \{x\}$ is closed.

Reconstructing continuous functions

We want to establish the following result.

Let $(X, \tau)$ and $(Y, \gamma)$ be topological spaces such that $\tau$ is $T_D$ . If $\varphi: \gamma \rightarrow \tau$ is a lattice isomorphism, then there is a continuous $f: X \rightarrow Y$ such that $\varphi = f^{-1}$ . And, if $\gamma$ is $T_0$ , then $f$ is unique.

We shall abuse the notation a little and consider $\varphi$ as a lattice isomorphism for the closed sets: $\varphi(A^c) = \varphi(A)^c$ .

Each $x$ uniquely determines the neighbourhood filter $\mathscr{F} = \tau(x)$ . Or $\mathscr{F}^c$ , or ${\mathscr{F}^c}^*$ or $\overline{\{x\}}$ . The isomorphism $\varphi$ associates those to corresponding filters, ideals and closed sets. From the fact that $X$ is $T_D$ , the set $\overline{\{x\}} \setminus \{x\}$ is closed. Let $E, F \subset Y$ be closed sets corresponding to $\overline{\{x\}} \setminus \{x\}$ and $\overline{\{x\}}$ .

Pick up an $y \in F \setminus E$ . Notice that there is no closed set between $E$ and $F$ . Not only that, but since $\varphi$ is a lattice isomorphism, $F$ is join-irreducible. And since $E$ and $F$ are closed, $F = E \cup \overline{\{y\}}$ is a union of closed sets. The join-irreducibility of $F$ implies that $F = \overline{\{y\}}$ . If $Y$ is $T_0$ , there is only one such $y$ . If not, simply pick any one (choice!) and define $f(x) = y$ . So that $f$ satisfies

$\begin{equation} \tag{😎} \label{closure_fx} \varphi\left(\overline{\{f(x)\}}\right) = \overline{\{x\}}. \end{equation}$

To see that $f^{-1} = \varphi$ — which implies that $f$ is continuous — notice that for a closed set $F \subset Y$ ,

$\begin{align} x \in f^{-1}(F) &\Leftrightarrow f(x) \in F \\ &\Leftrightarrow \overline{\{f(x)\}} \subset F \\ &\Leftrightarrow \varphi\left(\overline{\{f(x)\}}\right) \subset \varphi(F) \overset{\large\ref{closure_fx}}{\Leftrightarrow} \overline{\{x\}} \subset \varphi(F) \\ &\Leftrightarrow x \in \varphi(F). \end{align}$

Uniqueness

For any $x \in X$ , we have seen that there is a $y \in Y$ such that

$\begin{equation} \overline{\{x\}} = \varphi\left(\overline{\{y\}}\right). \end{equation}$

We have picked one such $y$ and defined $f(x) = y$ . This construction is the “only option”, because the continuity of $f$ implies that

$\begin{equation} \varphi\left(\overline{\{y\}}\right) = \overline{\{x\}} \subset f^{-1}\left(\overline{\{f(x)\}}\right) = \varphi\left(\overline{\{f(x)\}}\right). \end{equation}$

And, therefore,

$\begin{equation} \overline{\{y\}} \subset \overline{\{f(x)\}}. \end{equation}$

On the other hand, if $f^{-1} = \varphi$ ,

$\begin{align} f^{-1}\left(\overline{\{y\}}\right) &= \varphi\left(\overline{\{y\}}\right) = \overline{\{x\}} \\ &\Leftrightarrow f(x) \in \overline{\{y\}} \\ &\Rightarrow \overline{\{f(x)\}} \subset \overline{\{y\}}. \end{align}$

That is, the only options for $f(x)$ are those that satisfy $\overline{\{f(x)\}} = \overline{\{y\}}$ .

The uniqueness follows for the case where $Y$ is $T_0$ , because being $T_0$ is equivalent to

$\begin{equation} \overline{\{y\}} = \overline{\{z\}} \Rightarrow y = z. \end{equation}$

Conclusion

Every time I come up with some alternative I find simpler… I write the details down and realize that at the end, mine is not that simpler. Sometimes it is even more complicated. I do learn a lot writing it down, though. :-)

$T_D$ spaces

Definition

Reconstructing continuous functions

Uniqueness

Conclusion

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