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Let $X, Y$ be topological spaces and $f: X\to Y$ continuous. Show that $f$ maps connected sets onto connected sets.

I am not sure if I understand the definition of a connected set right. Our definitions are as follows:

A topological space $X$ is "connected" if $\emptyset$ and $X$ are the only sets, which are closed and open in $X$.

A subset of a topological space is "connected" if it is "connected" with regards to the subspace topology.

Therefor a set $A\subseteq X$ is connected, if $A$ and $\emptyset$ are the only closed and open ("clopen") sets with regards to the subspace topology?

Now for the proof:

Let $\emptyset\neq A\subset X$ be connected. Therefor $A$ is not clopen with regards to the subspace topology.

I have to show, that $f(A)$ is not clopen with regards to the subspace topology.

Suppose $\emptyset\neq f(A)\subset Y$ is clopen. Since $f$ is continuous the preimage of open and closed sets is open and closed. Hence $A$ is clopen. Which contradicts the assumption.

Is this correct? Thanks in advance.

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  • $\begingroup$ In your proof, you first assume that $A$ is a connected subset of $X$. It looks like you will try to prove that $f(A)$ is connected. Then conclude that $A$ is not clopen with respect to the subspace topology. But $A$ as a subspace is clopen. You say that your aim is to prove that $f(A)$ is not clopen with respect to the subspace topology, but again it is. $\endgroup$
    – user583185
    Commented Aug 13, 2018 at 10:57
  • $\begingroup$ So, I take $\emptyset\neq B\subset A$ which is not clopen, since $A$ is connected and I have to show, that $f(B)$ is not clopen? $\endgroup$
    – Cornman
    Commented Aug 13, 2018 at 10:57
  • $\begingroup$ @rTSlines My comment above is refering to your first comment. Was it wrong? $\endgroup$
    – Cornman
    Commented Aug 13, 2018 at 10:58
  • $\begingroup$ Well, to begin with it is not necessary to prove that $f$ maps all connected subsets of $X$ to connected. It is enough to assume that $X$ is connected and prove that $f(X)$ is connected. Since $X$ and $Y$ are arbitrary, with $X$ any connected space, then you are proving the required statement. $\endgroup$
    – user583185
    Commented Aug 13, 2018 at 10:58
  • $\begingroup$ So, what you need to show is that no proper subset of $f(X)$ is clopen. Assume $B\subset f(X)$ is clopen. Then look at $f^{-1}(B)\subset X$. $\endgroup$
    – user583185
    Commented Aug 13, 2018 at 11:00

1 Answer 1

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You are saying that $A\subseteq X$ is not clopen wrt subspace topology, but this statement is not true. Every subset $A$ of $X$ is clopen wrt the subspace topology on $A$ inherited from $X$.

Under the condition that $A$ is connected it must be proved that $f(A)$ is connected (not that $f(A)$ is not clopen).

This comes to the same as proving that no set $U\subset f(A)$ exists that is clopen wrt the subspace topology of $f(A)$ and satisfies $U\notin\{\varnothing,f(A)\}$.

Suppose that there is such a set.

Then its preimage under $f$ is a subset $V$ of $A$ that is clopen wrt the subspace topology of $A$ and satisfies $V\notin\{\varnothing, A\}$.

Proved is then: $f(A)$ not connected $\implies A$ not connected, or equivalently:

$$A\text{ connected }\implies f(A)\text{ connected}$$

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