The image of a connected set is connected

Stub grade: A*

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Doing some work while I've got a bit of time, majority written right before bed, so needs checking. This is also "the theorem" of connectedness

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

We do this proof by contrapositive, that is noting that: $(A\implies B)\iff((\neg B)\implies(\neg A))$, as such we will show:

• if $f(A)$ is disconnected then $A$ is disconnected

Let us begin:

• Suppose $f(A)$ is disconnected, and write $(f(A),\mathcal{K}_{f(A)})$ for the topological subspace on $f(A)$ of $(Y,\mathcal{ K })$.
  • Then there exists $U,V\in\mathcal{K}_{f(A)}$ such that $U$ and $V$ disconnect $f(A)$
    • Notice that $f^{-1}(U)$ and $f^{-1}(V)$ are disjoint
      • PROOF HERE
    • Notice also that $A\subseteq f^{-1}(U)\cup f^{-1}(V)$ (we may or may not have: $A=f^{-1}(U)\cup f^{-1}(V)$, as there might exist elements outside of $A$ that map into $f(A)$ notice)
      • PROOF HERE

Caution:We do not know whether or not $f^{-1}(U)$ or $f^{-1}(V)$ are open in $X$ (and they may form a strict superset of $A$ so cannot be open in $A$)

Grade: A

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Working on it

OLD WORKINGS

• OLD
  • WORK
    • $f^{-1}(U)$ 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