Mapping to Indiscrete Space is Continuous

Theorem

Let $T_1 = \struct {S_1, \tau_1}$ be any topological space.

Let $T_2 = \struct {S_2, \tau_2}$ be the indiscrete topological space on $S_2$.

Let $\phi: S_1 \to S_2$ be a mapping.


Then $\phi$ is continuous.


Proof

From the definition of continuous:

$U \in \tau_2 \implies \phi^{-1} \sqbrk U \in \tau_1$

The only elements of $\tau_2$ are $S_2$ and $\O$, from which:

$\phi^{-1} \sqbrk {S_2} = S_1 \in \tau_1$
$\phi^{-1} \sqbrk \O = \O \in \tau_1$

$\blacksquare$


Sources

  • 1975: W.A. Sutherland: Introduction to Metric and Topological Spaces ... (previous) ... (next): $3$: Continuity generalized: topological spaces: Exercise $3.9: 6$
  • 1978: Lynn Arthur Steen and J. Arthur Seebach, Jr.: Counterexamples in Topology (2nd ed.) ... (previous) ... (next): Part $\text {II}$: Counterexamples: $4$. Indiscrete Topology: $8$
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