# Proving that a continuous $f:X \to Y ; \ X,Y- \text{topological spaces}$ and $A \subseteq X ; A \text{ connected} \implies f(A) \text{ connected}$

Proving that $f:X \to Y ; \ X,Y- \text{topological spaces}$ and $A \subseteq X ; A \text{ connected} \implies f(A)-\text{connected}$

The answer is given like this just one step I do not understand and will highlight: It is proving using contraposition or $f(A) \text{ not connected} \implies A \text{ not connected}$, so having this in mind we have: $$f(A)= U \cup V;U,V \text{open sets }; U \cap V= \emptyset \\ f^{-1}(U), f^{-1}(V) \text{ are open in X.}\\$$

$$f^{-1}(U) \cap f^{-1}(V)= \emptyset \text{ because } f(A \cap B)\subseteq f(A) \cap f(B) \text{, so} \\ A \subseteq f^{-1}(U) \cup f^{-1}(V). \text{(not possible)}$$

I've looked at this problem many times, I might not be aware of some set properties maybe. I would greatly appreciate if someone could take apart these two lines in simple steps, because I am at a loss...

• Do you want any conditions on $f$? – preferred_anon Jul 23 '15 at 19:17
• Edited, continuous, thanks for the heads up – user246310 Jul 23 '15 at 19:18
• Who is the set B? – InsideOut Jul 23 '15 at 19:21
• that's like a general set rule the professor wrote supposedly – user246310 Jul 23 '15 at 19:22
• Maybe he wrote a general situation. – InsideOut Jul 23 '15 at 19:27

Here is a more detailed proof to make sure there are no loose ends.

Suppose that $f(A)$ is not connected. Then, there exist sets $U$ and $V$ such that:

• $U\subseteq f(A)$, $V\subseteq f(A)$;
• $U$ and $V$ are open with respect to the subspace topology on $f(A)$ inherited from $Y$;
• $U$ and $V$ are not empty;
• $U$ and $V$ are disjoint; and
• $U\cup V=f(A)$.

Now, given the definition of the subspace topology, there exist sets $\widehat{U}\subseteq Y$ and $\widehat{V}\subseteq Y$ that are open with respect to the “main” topology on $Y$ and

• $U=\widehat U\cap f(A)$;

• $V=\widehat V\cap f(A)$.

Next, observe that $f^{-1}(\widehat U)$ and $f^{-1}(\widehat V)$ are open subsets of $X$ because $f$ is continuous. Therefore, $f^{-1}(\widehat U)\cap A$ and $f^{-1}(\widehat V)\cap A$ are open subsets of $A$ with respect to the subspace topology on $A$ inherited from $X$. If one shows that

• $f^{-1}(\widehat U)\cap A$ and $f^{-1}(\widehat V)\cap A$ are not empty;

• $f^{-1}(\widehat U)\cap A$ and $f^{-1}(\widehat V)\cap A$ are disjoint; and

• $[f^{-1}(\widehat U)\cap A]\cup[f^{-1}(\widehat V)\cap A]=A$,

then one will have shown that $A$ is disconnected.

For the first claim, since $U$ is not empty, one can take $u\in U$. Then, $u\in\widehat U$ and $u\in f(A)$. In turn, there exists some $a\in A$ such that $u=f(a)$, which implies also that $a\in f^{-1}(\{u\})\subseteq f^{-1}(\widehat U)$. Therefore, $a\in f^{-1}(\widehat U)\cap A$. That $f^{-1}(\widehat V)\cap A$ is not empty can be shown similarly.

As for the second claim, suppose, for the sake of contradiction, that $a\in f^{-1}(\widehat U)\cap f^{-1}(\widehat V)\cap A$. Then, $$f(a)\in \widehat U\cap\widehat V\cap f(A)=[\widehat U\cap f(A)]\cap[\widehat V\cap f(A)]=U\cap V=\varnothing,$$ which is impossible. Hence, $f^{-1}(\widehat U)\cap A$ and $f^{-1}(\widehat V)\cap A$ are disjoint.

Finally, suppose that $a\in A$. Then, $f(a)\in f(A)=U\cup V$. If $f(a)\in U=\widehat U\cap f(A)$, then one has $f(a)\in\widehat U$, so $a\in f^{-1}(\widehat U)$. If $f(a)\in V$, then, analogously, $a\in f^{-1}(\widehat V)$. It follows that $$A\subseteq[f^{-1}(\widehat U)\cap A]\cup[f^{-1}(\widehat V)\cap A]$$ and the other inclusion is obvious. Hence, the third claim, too, holds, showing that $A$ is disconnected.

• @RaceBannon It is. The OP had specifically requested an explication of a terse, two-line proof. I assume they are learning topology right now—accordingly, I preferred erring on the side of being too verbose to omitting any step of the argument and leaving the OP still confused. – triple_sec Jul 23 '15 at 21:19

If $f(A) =U\cup V$, with $U\cap V=\emptyset$, then $$A\subseteq f^{-1}f(A)=f^{-1}(U\cup V)\subseteq f^{-1}(U)\cup f^{-1}(V)$$ and these are disjoint since $U\cap V=\emptyset$.

• You are welcome :) – InsideOut Jul 23 '15 at 19:31
• one question: Its true that the maps of two disjointed sets are disjointed in a general situation ? – user246310 Jul 23 '15 at 19:32
• @JerryWest, no, it isn't. Two disjoint sets can have non disjoint images through a non injective mapping. But the preimages of two disjoint sets are always disjoint no matter if the mapping is not an injection (be sure to check this - it has to do with the fact that functions are well defined, in the sense that they relate to the same element in their domain exactly one element in their counter-domain.) – Tarc Jul 24 '15 at 11:38

Maybe the right general set-theoretical laws to be used are:

• If $U$ and $V$ are disjoint sets, then $f^{-1}(U)$ and $f^{-1}(V)$ are also disjoint;
• $f^{-1}(U \cup V) = f^{-1}(U) \cup f^{-1}(V)$;
• $A \subseteq f^{-1}\big(f(A)\big)$.

These alone suffice to show that:

If $f(A) = U \cup V$, with disjoint $U$ and $V$, then $A \subseteq f^{-1}(U) \cup f^{-1}(V)$ with disjoint $f^{-1}(U)$ and $f^{-1}(V)$.

That's not hard to prove the following characterization:

Theorem. A topological space $X$ is connected if and only if every continuous map of $X$ into a discrete space is constant.

One can use this theorem to solve your problem as follows:

Let $g:f(A)\longrightarrow Y$ be a continuous map into a discrete space $Y$ where $f(A)$ is endowed with the subspace topology .

If $g$ weren't constant one could find $y_1, y_2\in f(A)$ such that $g(y_1)\neq g(y_2)$. In particular, $y_1\neq y_2$. But $y_1=f(x_1)$ and $y_2=f(x_2)$ for some $x_1, x_2\in A$. Also, $x_1\neq x_2$ (otherwise $y_1=y_2$). Hence, $gf:A\longrightarrow Y$ is a continuous map satisfying $gf(x_1)=g(y_1)\neq g(y_2)=gf(x_2)$, a contradiction.

Obs: To justify that $gf:A\longrightarrow Y$ is continuous you must use the following:

If $h:X\longrightarrow Y$ is a continuous map between topological spaces and $Z\subseteq Y$ is a subspace such that $h(X)\subseteq Z$ then $h:X\longrightarrow Z$ is continuous.