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I searched for "path connected connected" and didn't find anything that looked relevant on this site. I also Googled "path connected implies connected" and found a number of proofs that were more or less the same, showing that if we assume a path-connected set is not connected, then $[0, 1]$ is disconnected, which is a contradiction.

I have two questions. The first is whether my $\epsilon$-$\delta$ proof (shown below) that path-connectedness implies connectedness withstands scrutiny (specifically the section in bold). The second question is about trying to understand the proof that path-connectedness implies connectedness in the book I am studying (Elementary Analysis by Kenneth Ross, second edition). The proof in Ross' book is slightly different from the standard one described in the first paragraph above.

My notation is consistent with that from Ross' book throughout.

Theorem to be proven: if $E$ in metric space $(S, d)$ is path-connected, then $E$ is connected.

My proof: Assume that $E$ is disconnected, so that there are open sets $U_1, U_2 \subseteq S$ such that

  1. $(E \cap U_1) \cap (E \cap U_2) = \emptyset$,

  2. $(E \cap U_1) \cup (E \cap U_2) = E$,

  3. $E \cap U_1 \neq \emptyset$, and

  4. $E \cap U_2 \neq \emptyset$.

Consider $s \in E \cap U_1$ and $t \in E \cap U_2$. By path-connectedness there exists a continuous function $\gamma \colon [a, b] \to E$ such that $\gamma (a) = s$ and $\gamma (b) = t$.

Since $(E \cap U_1) \cap (E \cap U_2) = \emptyset$, there must be some minimum distance $\epsilon > 0$ such that for all $\alpha \in E \cap U_1$ and $\beta \in E \cap U_2$, we have $d(\alpha, \beta) \geq \epsilon$.

By continuity of $\gamma$, using $s, t$ and $\epsilon$ as above, there must exist a $\delta > 0$ such that

$\lvert b - a \rvert < \delta$ implies $d \big( \gamma(b), \gamma(a) \big) = d(t, s) < \epsilon$.

This contradicts the bolded section about minimum distance, whence we conclude that $E$ must be connected. $\square$

Question: is the assumption in bold about a minimum distance $\epsilon$ justifiable?

Proof by Ross: the setup is the same as in my proof. The divergence begins where Ross considers $F = \gamma \big( [a, b] \big)$. He claims that substituting $F$ for $E$ in conditions 1-4 above does not affect the veracity of those conditions. In other words, $F$ is disconnected. However, that contradicts the fact that $\gamma$, as a continuous function whose domain is connected (i.e. the interval $[a, b] \in \mathbb{R}$), must have a connected image $F = \gamma \big( [a, b] \big)$. Therefore we conclude that $E$ is in fact connected. $\square$

Question: How do I know that $F \cap U_1$ and $F \cap U_2$ satisfy conditions 1-4?

For condition 1, I can see that each bracketed term becomes no larger when substituting in $F$, so it makes sense that the intersection should still be the empty set. For condition 2, the best I can do is wave my hands and say "well, $U_1$ and $U_2$ 'cover' all of $E$, so they have to cover all of $F$, which is a subset of $E$."

For the third condition, I note that $\gamma (a) = s \in F$, and $s \in E \cap U_1$ by assumption, therefore $s \in F \cap U_1$ which proves that $F \cap U_1 \neq \emptyset$. Verification of the fourth condition is entirely similar.

Thanks for any help, and sorry for the long post.

RobPratt
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Novice
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1 Answers1

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The implication in bold is not correct. The fact that two subsets $A$, $B$ of a metric space are disjoint does not imply that there is an $\varepsilon >0 $ such that for all pairs $(a,b) \in A \times B$, $d(a,b) \ge \varepsilon $. Think for instance of the case $S=\mathbb{R}$, $A=(-1,0)$, $B=(0,1)$. Of course, you might say, this example does not work because $A \cup B$ is not path-connected. Well, of course it isn't: you cannot find an actual counterexample, since the theorem is true :) The point I want to make is just that the implication does not hold. (Tangentially, as you can see here, it would hold if $A$ was compact and $B$ was closed, but this is not assumed here).

I think you cannot get away without using the fact that $[0,1]$ is connected in some way, because this is the very heart of the definition of path-connectedness, and it is a necessary condition for path-connectedness to imply connectedness.

As for your second question: your argument for condition 2 is quite correct. You can just make it a bit more formal if you wish: if you assume that the union is not the whole of $F$, then there is a point in $F$ (hence in $E$, since $F$ is a subset of $E$) which lies in neither $U_1$ nor $U_2$. But this contradicts condition 2 for $E$.

57Jimmy
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  • I was thinking about something like letting $S = \mathbb{R}, E = (-1, 1), U_1 = (-1, 0)$ and $U_2 = (0, 1)$. I was getting hung up on the fact that I cannot use these sets to meet condition 2. I would have to close one of the endpoints, say $U_1^{\prime} = (-1, 0]$, but then I would fail condition 1. Of course I can see now that this line of thought is silly because $E$ is connected, so there are no sets that can disconnect it. To your point, I see that the sets $(-1, 0)$ and $(0, 1)$ make it clear that there is no minimum distance between them. – Novice May 05 '20 at 23:23
  • Regarding the second question, I see that if $(F \cap U_1) \cup (F \cap U_2) \neq F$, then that would imply $(F \cap U_1) \cup (F \cap U_2) \subset F$ (since the other way around is nonsensical). So assume there is a point $f \in F$, but $f \notin (F \cap U_1) \cup (F \cap U_2)$. Therefore, as you say we can deduce that $f \in E$ but $f \notin U_1$ and $f \notin U_2$. This would violate condition 2, and therefore we must have $(F \cap U_1) \cup (F \cap U_2) = F$. – Novice May 05 '20 at 23:35
  • Yes, that's precisely the argument – 57Jimmy May 05 '20 at 23:43