# Path connected implies connected

I am learning point set topology, and I want to be able to prove that a path connected set is also a connected set. I found many results by searching, but none of which I understand fully. One stands out from topospaces wiki:

Given: A path-connected topological space $$X$$

To prove: $$X$$ is connected.

Proof: We do this proof by contradiction. Suppose $$X$$ is not connected. Then, there exist nonempty disjoint open subsets $$U, V \subseteq X$$ such that $$X=U \cup V$$. Pick a point $$x \in U$$ and a point $$y \in V$$.

By assumption, there exists a continuous function $$f:[0,1] \rightarrow X$$ such that $$f(0)=x$$ and $$f(1)=y$$. $$\color{blue}{\text{Consider the subsets f^{-1}(U) and f^{-1}(V). These are disjoint in [0,1] and their union is [0,1]}}$$. By the continuity of $$f$$, they are both open in $$[0,1]$$. Finally, since $$0 \in f^{-1}(U)$$ and $$1 \in f^{-1}(V)$$, they are both nonempty. We have thus expressed $$[0,1]$$ as a union of two disjoint nonempty open subsets, a contradiction to the fact that $$[0,1]$$ is connected. This completes the proof.

I understand all parts of this proof except for the highlighted blue part. I do not know why the inverse images of two open disjoint sets in $$X$$ are also disjoint in $$[0,1]$$, and union to equal it. Is there some general property inverse images which means that inverse images of disjoint open sets are also disjoint? If so, why? (i.e. is there a proof?)

• Yes, if $f\colon A\rightarrow B$ is any function between sets (nothing to do with topology) and $M,N\subseteq B$, then $f^{-1}(M\cap N)=f^{-1}(M)\cap f^{-1}(N)$ and $f^{-1}(M\cup N)=f^{-1}(M)\cup f^{-1}(N)$. I highly encourage you to try and prove these identities yourself if you don't know them. Commented Sep 25, 2022 at 12:34
• @Thorgott I came across that while trying to find proofs, but I was unable to do a proof by expanding everything into its definition (e.g. $f^{-1}[B]=\{x \in X: f(x) \in B\}$). It made intuitive sense though, since an inverse image $f^{-1}[B]$ returns everything that the image $f$ could be input to return $B$. So if $B$ is the union of $C$ and $D$, it will return everything that could result in members from $C$ or $D$. Which is the same as returning everything that could result in members from $C$, in a union with everything that could result in members from $D$. How could it be anything else? Commented Sep 25, 2022 at 12:54
• The reason why I dropped that line of working toward a proof though, was how could I know that the sets where disjoint? However I have just realised that the inverse image of the empty set is the empty set, which allows me to complete the proof provided I have one for the result you suggest I work on. Commented Sep 25, 2022 at 12:54
• If $x\in f^{-1}(U)\cap f^{-1}(V)$ then $f(x)\in U$ and $f(x)\in V$, contradicting that $U,V$ are disjoint. For all $x$, $f(x)\in X$ so either $f(x)\in U$ or $f(x)\in V$. Then either $x\in f^{-1}(U)$ or $x\in f^{-1}(V)$. Commented Sep 25, 2022 at 14:33

$$\color{blue}{\text{Consider the subsets f^{-1}(U) and f^{-1}(V). These are disjoint in [0,1] and their union is [0,1]}}$$

We prove the text in blue, where $$U$$ and $$V$$ are disjoint, $$X=U\cup V$$, and $$f$$ is a continuous function $$f:[0,1] \rightarrow X$$ such that $$f(0)=x$$ and $$f(1)=y$$.

First we show that $$f^{-1}(A\cup B)=f^{-1}(A)\cup f^{-1}(B)$$, and $$f^{-1}(A\cap B)=f^{-1}(A)\cap f^{-1}(B)$$, as per a comment by Thorgott. We do this using the definitions of inverse images, set union, set intersection, in set builder notation.

• The union of two sets $$A\cup B$$ is $$\{x|x\in A\text{ or }x\in B\}$$.

• The intersection of two sets $$A\cap B$$ is $$\{x|x\in A\text{ and }x\in B\}$$.

• The inverse image $$f^{-1}(A)$$ is defined to be $$\{x\in X|f(x)\in A\}$$. Inverse images are defined for all functions, invertible or otherwise.

We then have:

\begin{align} z\in f^{-1}(A\cup B) &\iff f(z)\in A\cup B\\ &\iff f(z)\in A\text{ or }f(z)\in B\\ &\iff z\in f^{-1}(A)\text{ or }z\in f^{-1}(B)\\ &\iff z\in f^{-1}(A)\cup f^{-1}(B)\\ \\ \therefore~f^{-1}(A\cup B)=&f^{-1}(A)\cup f^{-1}(B) \end{align}

\begin{align} z\in f^{-1}(A\cap B) &\iff f(z)\in A\cap B\\ &\iff f(z)\in A\text{ and }f(z)\in B\\ &\iff z\in f^{-1}(A)\text{ and }z\in f^{-1}(B)\\ &\iff z\in f^{-1}(A)\cap f^{-1}(B)\\ \\ \therefore~f^{-1}(A\cap B)=&f^{-1}(A)\cap f^{-1}(B) \end{align}

It remains to be shown that the subsets $$f^{-1}(U)$$ and $$f^{-1}(V)$$. These are disjoint in $$[0,1]$$ and their union is $$[0,1]$$. We use the fact that $$U$$ and $$V$$ are disjoint, and that $$f$$ maps $$[0,1]$$ to $$X=U\cup V$$ to demonstrate this.

\begin{align} f^{-1}(U)\cup f^{-1}(V)&=f^{-1}(U\cup V)\\ &=f^{-1}(X)\\ &=[0,1] \end{align}

\begin{align} f^{-1}(U)\cap f^{-1}(V)&=f^{-1}(U\cap V)\\ &=f^{-1}(\emptyset)\\ &=\{x\in [0,1]|f(x)\in \emptyset\}\\ &=\emptyset\\ \end{align}

$$\therefore$$ $$f^{-1}(U)$$ and $$f^{-1}(V)$$ are disjoint, and their union is equal to $$[0,1]$$.