The image of a connected set is connected

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Doing some work while I've got a bit of time

Caution:This is being done RIGHT BEFORE BED - do not rely on it until I've checked it

Statement

Let $(X,\mathcal{ J })$ and $(Y,\mathcal{ K })$ be topological spaces and let $f:X\rightarrow Y$ be a continuous map. Then, for any $A\in\mathcal{P}(X)$, we have:

• If $A$ is a connected subset of $(X,\mathcal{ J })$ then $f(A)$ is connected subset in $(Y,\mathcal{ K })$

Proof

Suppose $f(A)$ is disconnected, and $(f(A),\mathcal{K}_{f(A)})$ is a topological subspace of $(Y,\mathcal{ K })$.

• Then there exist $U,V\in\mathcal{K}_{f(A)}$ such that $U$ and $V$ disconnect $f(A)$
  • $f^{-1}(U)$ and $f^{-1}(V)$ are disjoint by $f$ being a function and their union contains $A$ (but could be bigger than it, as we might not have $f^{-1}(f(A))=A$ of course!)
  • We apply the right-hand part of:
    • a subset of a topological space is disconnected if and only if it can be covered by two non-empty-in-the-subset and disjoint-in-the-subset sets that are open in the space itself

This completes the proof Caution:Good night, still to do, put page in the right place!