Cfrgtkky

Take the definition of a connected topological space to be one that has no clopen sets other than the space itself or the empty set.

I claim to prove that if $f: X to Y$ is a continuous function between topological spaces and that if $X$ is connected then the image $f(X)$ is connected also.

proof:

Assume for a contradiction that $X$ is connected and $f(X)$ is disconnected.

Then $exists U subset f(X)$ such that $U$ is clopen under the subset topology of $f(X)$ and $ U neq emptyset$, $U neq f(X)$.

Since $f$ is continuous, $f^{-1}(U)$ is both open and closed since $U$ is clopen. Hence $f^{-1}(U)$ is a clopen set in $X$, all that is left to do is to show that it is non trivial.

Since by assumption $U neq emptyset$ and $U subset f(X)$; $f^{-1}(U) neq emptyset$.

Furthermore $U neq f(X)$ so $exists a in f(X)backslash U$

$therefore f^{-1}{a} in X backslash f^{-1}(U)$

$therefore f^{-1}(U) neq X$

So $f^{-1}(U)$ is a non trivial clopen subset of $X$ and $X$ is disconnected contradictory to what was assumed. Therefore $f(X)$ must be connected also. $square$

Is this proof correct? It is different from the one I have seen in my topology class and I am dubious of its simplicity.

You're almost good;

You cannot state that $f^{-1}(U)$ is clopen because you don't know that $f(X)$ is open or closed in $Y$!

To overcome this difficulty simply notice that the restriction map $f:X to f(X)$ is itself continuous, when $f$ is. Then the proof becomes quite trivial. Any separation of $f(X)$ projects back to two non empty sets, whose union must be $X$..