Continuous function maps connected set onto connected set

Question

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.

Answer 1

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:

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