T1/2 Space is T0 Space

From ProofWiki
Jump to navigation Jump to search

Theorem

Let $T = \left({S, \tau}\right)$ be a $T_{\frac 1 2}$ topological space.

Then $T$ is $T_0$ space.


Proof

By Characterization of T0 Space by Closures of Singletons it suffices to prove that

$\forall x, y \in S: x \ne y \implies x \notin \left\{{y}\right\}^- \lor y \notin \left\{{x}\right\}^-$

where $\left\{{y}\right\}^-$ denotes the closure of $\left\{{y}\right\}$

Let $x, y$ be points of $T$ such that

$x \ne y$

Aiming for a contradiction suppose

$x \in \left\{{y}\right\}^- \land y \in \left\{{x}\right\}^-$

We will prove that

$x \notin \left\{{x}\right\}'$

where $ \left\{{x}\right\}'$ denotes the derivative of $\left\{{x}\right\}$

Aiming for a contradiction suppose

$x \in \left\{{x}\right\}'$

As $S$ is open by definition of topological space by Characterization of Derivative by Open Sets:

$\exists z \in S: z \in \left\{{x}\right\} \cap S \land z \ne x$

Bu definition of intersection:

$z \in \left\{{x}\right\}$

This contradicts $z \ne x$ by definition of singleton

Thus $x \notin \left\{{x}\right\}'$


We will prove that:

$(1): \quad \lnot \forall G \in \tau: y \in G \implies \left\{{x}\right\} \cap G \ne \varnothing$

Aiming for a contradiction suppose

$\forall G \in \tau: y \in G \implies \left\{{x}\right\} \cap G \ne \varnothing$

As sublemma we will show that

$\forall U \in \tau: y \in U \implies \exists r \in S: r \in \left\{{x}\right\} \cap U \land y \ne r$

Let $U \in \tau$ such that

$y \in U$

Then by assumption

$\left\{{x}\right\} \cap U \ne \varnothing$

By definition of empty set:

$\exists z: z \in \left\{{x}\right\} \cap U$

By definition of intersection:

$z \in \left\{{x}\right\}$

Then by definition of singleton:

$z = x \ne y$

Thus:

$\exists r \in S: r \in \left\{{x}\right\} \cap U \land y \ne r$


Then by Characterization of Derivative by Open Sets:

$y \in \left\{{x}\right\}'$

By definition of relative complement:

$y \notin \complement_S \left({\left\{{x}\right\}'}\right) \land x \in \complement_S \left({\left\{{x}\right\}'}\right)$

By definition of $T_{\frac 1 2}$ space:

$\left\{{x}\right\}'$ is closed

By definition of closed set:

$\complement_S \left({\left\{{x}\right\}'}\right)$ is open

By $x \in \left\{{y}\right\}^-$ and by Condition for Point being in Closure:

$\left\{{y}\right\} \cap \complement_S \left({\left\{{x}\right\}'}\right) \ne \varnothing$

Then by definition of empty set:

$\exists z: z \in \left\{{y}\right\} \cap \complement_S \left({\left\{{x}\right\}'}\right)$

By definition of intersection:

$z \in \left\{{y}\right\} \land z \in \complement_S \left({\left\{{x}\right\}'}\right)$

By definition of singleton:

$z = y$

Thus $z \in \complement_S \left({\left\{{x}\right\}'}\right)$ contradicts $y \notin \complement_S \left({\left\{{x}\right\}'}\right)$

This ends the proof of $(1)$.


Then by $(1)$ ans Condition for Point being in Closure

$y \notin \left\{{x}\right\}^-$

This contradicts $y \in \left\{{x}\right\}^-$

$\blacksquare$


Sources