Let $X$ be a topological space and $\mathcal{A}= \{f: X \to X \mid f \text{ continuous}\}$. Determine the topology that $\mathcal{A}$ induces on $X$.

Let $X$ be a topological space and $\mathcal{A}= \{f: X \to X \mid f \text{ continuous}\}$. Determine the topology that $\mathcal{A}$ induces on $X$.

The definition I found for the induced topology by collection of maps is that it is the collection $\{f^{-1}_i (V) \mid V \subset X \}$ where $V$ is an open set. If each $f_i$ is continuous then for any $V$ open in $X$ we have that $f^{-1}_i (V)$ is open. And as $$f^{-1}_i (V)=\{x \in X \mid f(x) \in V\}$$ I think we have that the singletons are open which would imply that the induced topology is the discrete one, but I don't know if this is true?

Solution 1:

The induced topology $\tau_i$ by $\mathcal A$ has subbase

$$\mathcal S=\{f^{-1}[O]; f \in \mathcal{A}; O \in \tau_X\}$$

by definition.

If $U=f^{-1}[O] \in \mathcal{S}$ then $U \in \tau_X$ because $f$ is continuous on $X$ (as $f \in \mathcal A$). So it follows that $\tau_i \subseteq \tau_X$.

OTOH, if $U \in \tau_X$, then $1_X^{-1}[U]=U \in \mathcal S \subseteq \tau_i$ because the identity $1_X \in \mathcal A$, always. Hence $\tau_X \subseteq \tau_i$.

Conclusion: the induced topology is precisely the topology on $X$ that we started out with.