Editing Base (topology) (section)

==Weight and character==
We shall work with notions established in {{harv|Engelking|1989|loc=p. 12, pp. 127-128}}.

Fix <math>X</math> a topological space. Here, a '''network''' is a family <math>\mathcal{N}</math> of sets, for which, for all points <math>x</math> and open neighbourhoods ''U'' containing <math>x</math>, there exists <math>B</math> in <math>\mathcal{N}</math> for which <math>x \in B \subseteq U.</math> Note that, unlike a basis, the sets in a network need not be open.

We define the '''weight''', <math>w(X)</math>, as the minimum cardinality of a basis; we define the '''network weight''', <math>nw(X)</math>, as the minimum cardinality of a network; the '''character of a point''', <math>\chi(x,X),</math> as the minimum cardinality of a neighbourhood basis for <math>x</math> in <math>X</math>; and the '''character''' of <math>X</math> to be
<math display=block>\chi(X)\triangleq\sup\{\chi(x,X):x\in X\}.</math>

The point of computing the character and weight is to be able to tell what sort of bases and local bases can exist. We have the following facts:

* <math>nw(X) \leq w(X)</math>.
* if <math>X</math> is discrete, then <math>w(X) = nw(X) = |X|</math>.
* if <math>X</math> is Hausdorff, then <math>nw(X)</math> is finite if and only if <math>X</math> is finite discrete.
* if <math>B</math> is a basis of <math>X</math> then there is a basis <math>B'\subseteq B</math> of size <math>|B'|\leq w(X).</math>
* if <math>N</math> is a neighbourhood basis for <math>x</math> in <math>X</math> then there is a neighbourhood basis <math>N'\subseteq N</math> of size <math>|N'|\leq \chi(x,X).</math>
* if <math>f : X \to Y</math> is a continuous surjection, then <math>nw(Y) \leq w(X)</math>. (Simply consider the <math>Y</math>-network <math>fB \triangleq \{f(U) : U \in B\}</math> for each basis <math>B</math> of <math>X</math>.)
* if <math>(X,\tau)</math> is Hausdorff, then there exists a weaker Hausdorff topology <math>(X,\tau')</math> so that <math>w(X,\tau')\leq nw(X,\tau).</math> So ''a fortiori'', if <math>X</math> is also compact, then such topologies coincide and hence we have, combined with the first fact, <math>nw(X) = w(X)</math>.
* if <math>f : X \to Y</math> a continuous surjective map from a compact metrizable space to an Hausdorff space, then <math>Y</math> is compact metrizable.

The last fact follows from <math>f(X)</math> being compact Hausdorff, and hence <math>nw(f(X))=w(f(X))\leq w(X)\leq\aleph_0</math> (since compact metrizable spaces are necessarily second countable); as well as the fact that compact Hausdorff spaces are metrizable exactly in case they are second countable. (An application of this, for instance, is that every path in a Hausdorff space is compact metrizable.)

===Increasing chains of open sets===
Using the above notation, suppose that <math> w(X) \leq \kappa </math> some infinite cardinal. Then there does not exist a strictly increasing sequence of open sets (equivalently strictly decreasing sequence of closed sets) of length <math> \leq \kappa^+\!</math>.

To see this (without the axiom of choice), fix
<math display=block>\left \{ U_{\xi} \right \}_{\xi\in\kappa},</math>
as a basis of open sets. And suppose ''per contra'', that
<math display=block>\left \{ V_{\xi}\right \}_{\xi\in\kappa^{+}}</math>
were a strictly increasing sequence of open sets. This means
<math display=block>\forall \alpha<\kappa^+\!: \qquad V_{\alpha}\setminus\bigcup_{\xi<\alpha} V_{\xi} \neq \varnothing.</math>

For
<math display=block>x\in V_{\alpha}\setminus\bigcup_{\xi<\alpha}V_{\xi},</math>
we may use the basis to find some <math>U_\gamma</math> with <math>x</math> in <math>U_\gamma \subseteq V_\alpha</math>. In this way we may well-define a map, <math>f : \kappa^+ \!\to \kappa</math> mapping each <math>\alpha</math> to the least <math>\gamma</math> for which <math>U_\gamma \subseteq V_\alpha</math> and meets
<math display=block>V_{\alpha} \setminus \bigcup_{\xi<\alpha} V_{\xi}.</math>

This map is injective, otherwise there would be <math>\alpha < \beta</math> with <math>f(\alpha) = f(\beta) = \gamma</math>, which would further imply <math>U_\gamma \subseteq V_\alpha </math> but also meets
<math display=block>V_{\beta} \setminus \bigcup_{\xi<\alpha} V_{\xi} \subseteq V_{\beta} \setminus V_{\alpha},</math>
which is a contradiction. But this would go to show that <math>\kappa^+ \!\leq \kappa</math>, a contradiction.