Uncountable Fort Space is not First-Countable

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Theorem

Let $T = \struct {S, \tau_p}$ be a Fort space on an uncountable set $S$.


Then $T$ is not a first-countable space.


Proof

Let $\UU$ be a countable set of open neighborhoods of $p$.

Let $U \in \UU$.

Then as $p \in U$, $p \notin \relcomp S U$ and so for $U$ to be open it must follow that $\relcomp S U$ is finite.

From De Morgan's Laws: Complement of Intersection we have:

$\displaystyle H := \bigcup_{U \mathop \in \UU} \relcomp S U = \relcomp S {\bigcap \UU}$

From Countable Union of Countable Sets is Countable it follows that $H$ can be no more than countable.

So $\displaystyle \bigcap \UU = \relcomp S H$ must be uncountable.

So (trivially):

$H \ne \set p$

So:

$\exists q \ne p: q \in \displaystyle \bigcap \UU$

So $\relcomp S {\set q}$ is an open neighborhood of $P$ which does not contain any of the elements of $\UU$.

So $\UU$ is not a countable local basis of $p$.

Hence by definition $T$ is not a first-countable space.

$\blacksquare$


Sources