The closure of a continuous image of a closure is the closure of the image

up:: Topologically Continuous Function

Continuous functions map close inputs to close outputs.

Given a Topologically Continuous function $f:X\to Y$, it can be proven that, for any set $D\subseteq X$,

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

$(\subseteq )$

Using that A continuous function’s image of the closure is a subset of the closure of its image for a Topologically Continuous Function $f:(X,{\tau }_X)\to (Y,{\tau }_Y)$, one knows that, for any subset $D\subseteq X$,

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

From this, since Closure preserves subset ordering, we have that

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

$(\supseteq )$

On the other hand, since we know that All sets are contained inside their closure, and that Function images preserve subset ordering, we have that

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

Since Closure preserves subset ordering, we also have that

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

Conclusion

Thus, for all continuous functions,

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

Operators on the Power Set

We can see the image of $f$ and the closure $Cl$ as operators over power sets, as follows

$$\{\begin{array}{l}I{m}_f:\mathbb{P}(X)\to \mathbb{P}(Y)\\ C{l}_X:\mathbb{P}(X)\to \mathbb{P}(X)\\ C{l}_Y:\mathbb{P}(Y)\to \mathbb{P}(Y)\end{array}$$

Using that, it can be observed that

$$C{l}_Y\circ I{m}_f\circ C{l}_X=C{l}_Y\circ I{m}_f\circ 1_{\mathbb{P}(X)}$$

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References

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