The image of a connected set is connected

Stub grade: A*

This page is a stub

This page is a stub, so it contains little or minimal information and is on a to-do list for being expanded.The message provided is:

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 [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:
    • 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!