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
Contents
Statement
Let [ilmath](X,\mathcal{ J })[/ilmath] and [ilmath](Y,\mathcal{ K })[/ilmath] be topological spaces and let [ilmath]f:X\rightarrow Y[/ilmath] be a continuous map. Then, for any [ilmath]A\in\mathcal{P}(X)[/ilmath], we have:
- If [ilmath]A[/ilmath] is a connected subset of [ilmath](X,\mathcal{ J })[/ilmath] then [ilmath]f(A)[/ilmath] is connected subset in [ilmath](Y,\mathcal{ K })[/ilmath]
Proof
Suppose [ilmath]f(A)[/ilmath] is disconnected, and [ilmath](f(A),\mathcal{K}_{f(A)})[/ilmath] is a topological subspace of [ilmath](Y,\mathcal{ K })[/ilmath].
- Then there exist [ilmath]U,V\in\mathcal{K}_{f(A)} [/ilmath] such that [ilmath]U[/ilmath] and [ilmath]V[/ilmath] disconnect [ilmath]f(A)[/ilmath]
- [ilmath]f^{-1}(U)[/ilmath] and [ilmath]f^{-1}(V)[/ilmath] are disjoint by [ilmath]f[/ilmath] being a function and their union contains [ilmath]A[/ilmath] (but could be bigger than it, as we might not have [ilmath]f^{-1}(f(A))=A[/ilmath] of course!)
- We apply the right-hand part of:
This completes the proof Caution:Good night, still to do, put page in the right place!