Identity Mapping to Coarser Topology is Continuous
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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
- $\tau_2$ 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 $\tau_1$.
$\blacksquare$