# The image of a connected set is connected

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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 [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

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

• if [ilmath]f(A)[/ilmath] is disconnected then [ilmath]A[/ilmath] is disconnected

Let us begin:

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

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

Grade: A
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Working on it

### OLD WORKINGS

This completes the proof