Interior covering and path connectedness

Question

If $X_1, X_2$ are subsets of $X$ such that $X = int \, X_1 \cup \, int \, X_2$. Prove that, if $X$ is path connected, then each path component of $X_1$ meets $X_2$.

I would be really appreciated if a hint is given. I tried proof by contradiction.

My attempt:

• The special case when either of the interiors is empty implies $X_i=X$ hence statement is trivially true.

• Let $x \in X_1$ and $C_x$ be the path component in $X_1$. Suppose $C_x \cap X_2=\emptyset$, let $y \in int \, X_2$, there is a path $b:x \rightarrow y$ in $X$.

• It suffices to show that there are two disjoint open sets that cover the domain of $b$ (which is an interval). I believe one candidate is $b^{-1}(int \, X_2)$. The other I am stuck.

Answer 1

The meaning of the phrase "sets $A, B$ meet" is that $A \cap B \ne \emptyset$.

We distinguish various cases.

Case 1: $X_1 = \emptyset$. Then the assertion is wrong because $\emptyset$ is a path component of $X_1$ which does not meet $X_2$. However, note that it is a matter of taste if $\emptyset$ is regarded as path-connected or not. Some authors do not, but the usual definition says it is (so that $\emptyset$ has a single path component, namely $\emptyset$ itself). I adopted this point of view, but if you don't like to do so the assertion becomes trivially true because then $\emptyset$ does not have any path component.

Case 2: $X_1 \ne \emptyset$.

Case 2.1: $X_2 = \emptyset$. Then the assertion is wrong.

Case 2.2: $X_2 \ne \emptyset$.

Case 2.2.1: $int X_2 = \emptyset$. Then the assertion is trivially true because this implies $X_1 = X$ (as pointed out in the question).

Case 2.2.2: $int X_2 \ne \emptyset$. This is the only non-trivial case.

To prove the assertion we have to show that for each $x \in X_1$ there exists a path $v : [0,a] \to X$ such that $v([0,a]) \subset X_1$, $v(0) = x$ and $v(a) \in X_2$.

Case a: $x \notin int X_1$. Then $x \in int X_2$ and we can take any constant path $v(t) \equiv x$.

Case b: $x \in int X_1$. Choose $y \in int X_2$. Since $X$ is path-connected, there exists a path $u : [0,b] \to X$ such that $u(0) = x, u(b) = y$. If $u([0,b]) \subset int X_1$, we are ready. Otherwise define $c = inf \lbrace t \in [0,b] \mid u(t) \notin int X_1 \rbrace$. We have $c > 0$ and $u([0,c)) \subset int X_1$. $u(c)$ cannot be contained in $int X_1$ (otherwise a neighborhood of $c$ would be mapped into $int X_1$ which would contradict the definition if $c$). Hence $u(c) \in int X_2$ and we conclude that a neighborhood of $c$ is mapped into $int X_2$. But then we find $a \in (0,c)$ such that $u(a) \in int X_2$. This means that $v = u \mid_{[0,a]}$ is a path as required.

Answer 2

Note: this proof is broken in its second paragraph, as pointed out in the comments. I've chosen to leave this answer up since the approach seems promising.

I've found a direct proof of this claim. It's a little bit technical, but the key idea here is to use the path-connectivity of $X$ to create a path, then leverage our assumptions to find the first place that this path crosses into a path-component of $X_2$.

Let $x \in X_1$ and $y \in X_2$. Then, since $X$ is path connected, we can find a path $\gamma$ starting at $x$ and ending at $y$. We now leverage our assumption that $X$ is formed as the union of the interiors of $X_1$ and $X_2$. Let $\{V_i\}$ be the collection of the interiors of path-components of $X_1$ and $X_2$. Each $V_i$ is open. By assumption, $\cup_i V_i = X$. The collection $\{\gamma^{-1}(V_i)\}$ is then an open cover of $[0,1]$ by the continuity of $\gamma$. Since $[0,1]$ is compact, $\{\gamma^{-1}(V_i)\}$ has a finite subcover, $\mathscr{C} = \{\gamma^{-1}(V_1), ... , \gamma^{-1}(V_n)\}$.

We now place a nice order on $\mathscr{C}$. In particular, order by $\mathscr{C}$ by the greatest lower bounds of elements of each $\gamma^{-1}(V_i)$, where sets with the lower greatest lower bounds come first. Since path-components are disjoint, we are guaranteed that the preimages of distinct path-components of $X_1$ (or $X_2$) will have distinct greatest lower bounds. If the preimages of path-components of $X_1$ and $X_2$ happen to have the same greatest lower bound, place the preimage of the path-component of $X_1$ first.

With this ordering, $0 \in \gamma^{-1}(V_1)$. $V_1$ is either the interior of a path-component of $X_1$ or $X_2$. If $V_1$ is the interior of a component of $X_2$, then $\gamma(0) = x$ lies in $X_2$, hence $X_1 \cap X_2 \neq \emptyset$ and we're done. If $V_1$ is the interior of a path-component of $X_1$, consider $V_2$. (If there is no second element in the cover, then $\gamma(1) = y$ lies in $V_1$, and we're done.) Since $\mathscr{C}$ is a finite open cover of $[0,1]$, the ordering we gave $\mathscr{C}$ gaurantees that $\gamma^{-1}(V_1) \cap \gamma^{-1}(V_2)$ is nonempty. Since path-components of a space are disjoint, this shows that $V_2$ is a path-component of $X_2$, and the nonemptiness of $\gamma^{-1}(V_1) \cap \gamma^{-1}(V_2)$ yields a point $a \in [0,1]$ with $\gamma(a) \in V_1 \cap V_2$. Hence, $V_1$ and $V_2$ have nontrivial intersection. Since the interior of a set is contained in the set, this shows that the path components whose interiors are $V_1$ and $V_2$ have nonempty intersection.

In either case, we have shown that the path-component of $X_1$ containing $x$ has nontrivial intersection with a path-component of $X_2$. This completes the proof.