A continuous function's image of the closure is a subset of the closure of its image

up:: Topologically Continuous Function

Given a Topologically Continuous Function $f:(X,{\tau }_X)\to (Y,{\tau }_Y)$, it can be proven that Continuous functions premap closed sets to closed sets is equivalent to the assertion that, for each $D\subseteq X$, the function satisfies that

$$f(\overline{D})\subseteq \overline{f(D)}$$

”Thesis” implies usual definition

Assuming that $f(\overline{D})\subseteq \overline{f(D)}$, we seek to prove that $f^{-1}(F)\in \mathcal{F}({\tau }_X)$, for any closed set $F\in \mathcal{F}({\tau }_Y)$.

Let $F\in \mathcal{F}({\tau }_Y)$ a closed set. Considering $f^{-1}(F)\subseteq X$, per hypothesis, $f(\overline{f^{-1}(F)})\subseteq \overline{f(f^{-1}(F))}$.

Using The image of the preimage of a set is contained in the set ($f(f^{-1}(F))\subseteq F$) and that Closure preserves subset ordering yields

$$f(\overline{f^{-1}(F)})=\overline{f(f^{-1}(F))}\subseteq \overline{F}=F$$

Taking the preimage on both sides (using that Preimages preserve subset ordering), since The preimage of the image of a set contains the set $f^{-1}(F)\subseteq f^{-1}(f(f^{-1}(F)))$, we have that

$$\overline{f^{-1}(F)}\subseteq f^{-1}(F)$$

Since All sets are contained inside their closure, it is proved that $f^{-1}(D)=\overline{f^{-1}(D)}$, thus $f^{-1}(D)$ is closed, and therefore $f$ premaps closed sets (in $Y$) to closed sets (in $X$).

Usual definition implies “thesis”

Let $f$ be continuous in the usual sense, i.e. premap closed sets in $Y$ to closed sets in $X$.

Let $D\subseteq X$ be some arbitrary set. Since The preimage of the image of a set contains the set and that All sets are contained inside their closure, we have that

$$D\subseteq f^{-1}(f(D))\subseteq f^{-1}(\overline{f(D)})$$

Since $f^{-1}(F)\in \mathcal{F}({\tau }_X)$ for any closed set in $Y$ per hypothesis, and using that Closure preserves subset ordering, then we can take the Closure on both sides, yielding

$$\overline{D}\subseteq f^{-1}(\overline{f(D)})$$

Using that Function images preserve subset ordering, we have that

$$f(\overline{D})\subseteq f\left(f^{-1}\left(\overline{f(D)}\right)\right)$$

Then, using that The image of the preimage of a set is contained in the set for the set $\overline{f(D)}$, we have

$$f(\overline{D})\subseteq f\left(f^{-1}\left(\overline{f(D)}\right)\right)\subseteq \overline{f(D)}$$

Thus, all Topologically Continuous Functions (in the usual definition) satisfy this property.

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References

• Notas para um Curso de Física-Matemática, João Carlos Alves Barata. Cap. 30