Identity Mapping to Coarser Topology is Continuous

From ProofWiki
Jump to navigation Jump to search

Theorem

Let $S$ be a set.

Let $\tau_1$ and $\tau_2$ be topologies on $S$.

That is, let $T_1 = \struct {S, \tau_1}$ and $T_2 = \struct {S, \tau_2}$ be topological spaces.

Let $I_S: \struct {S, \tau_1} \to \struct {S, \tau_2}$ denote the identity mapping on $S$:

$\forall x \in S: \map {I_S} x = x$


Then:

$I_S: T_1 \to T_2$ is a continuous mapping

if and only if:

$\tau_2$ is coarser than $\tau_1$.


Proof

\(\ds \forall U \in \tau_2: \, \) \(\ds I_S^{-1} \sqbrk U\) \(\in\) \(\ds \tau_1\) Definition of Continuous Mapping (Topology)
\(\ds \leadstoandfrom \ \ \) \(\ds \forall U \in \tau_2: \, \) \(\ds U\) \(\in\) \(\ds \tau_1\) Definition of Identity Mapping
\(\ds \leadstoandfrom \ \ \) \(\ds \tau_2\) \(\subseteq\) \(\ds \tau_1\)

But $\tau_2 \subseteq \tau_1$ is the definition of $\tau_2$ being coarser than $\tau_1$.

$\blacksquare$