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Is there a general rule of what kind of sets it is easier to prove connectedness using path connectedness or regular connectedness?

I understand that path connected $\implies$ connected, but are there situations where it's easier to prove using path connectedness vs. connectedness and vice-versa? One example I know of is that proving that an interval is connected in $\mathbb{R}$ by constructing a function

$\gamma:[0,1]\rightarrow [a,b]: \gamma(t)=c+t(d-c), t\in [0,1]$

which is clearly continuous. The connectedness approach is a lot more laborious. So my question is, is there a general rule of what kind of sets it is easier to prove connectedness using path connectedness or regular connectedness?

Emir
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  • Given that connected doesn't imply path connected, I guess you're asking for what spaces these properties are equivalent? – Lepidopterist Mar 09 '12 at 03:54
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    Couldn't you just prove $[a,b]$ is connected by the same proof that $[0,1]$ is connected? You need to prove $[0,1]$ is connected anyhow to say that path-connected implies connected.

    Perhaps the most important class of spaces for which connected implies path-connected is manifolds.

     – Brett Frankel Mar 09 '12 at 04:01
  • I can use the fact that path connected implies connected without proof in my class. So I was wondering of what were some examples where this might come in handy -- where I am asked to prove some set is connected and use path connectedness instead – Emir Mar 09 '12 at 05:05
  • @jay: No, Emir is asking whether there is asking whether there is a good heuristic for deciding which of the two is easier to prove for spaces that are path-connected. – Brian M. Scott Mar 09 '12 at 09:05
  • To emphasize what Brett Frankel wrote, considering Emir's reply to it, being able to "use the fact that path connected implies connected without proof in my class" leads to an error of circular reasoning: you can't prove $[0,1]$ (or any other closed interval $[a,b]$ with $a < b$) is connected by proving it's path connected because the proof that path connectedness implies connectedness relies on knowing $[0,1]$ is connected. I've never seen a proof that path connectedness implies connectedness that doesn't depend on connectedness of $[0,1]$: it is the nontrivial input into the proof. – KCd Jan 18 '20 at 14:02

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In my experiences, in metric spaces, path connectedness is usually-not always,but usually- easier and gives connectedness easily. In general topological spaces,it's sometimes not so easy to construct paths and the general definition of connectedness is usually more useful.

Interestingly, there are cases of topological spaces where it is fairly easy to prove the space is connected,but difficult to prove it is path connected. For example,the topologist's sine curve is connected but not path connected. How do we derive path connectedness given connectedness? The following theorem answers the question:

Theorem: Let S be a subset of a topological space X. Then if S is connected and locally path connected, then S is path connected.

The proof is somewhat lengthy,but not difficult. (Hint to get you started: Consider a path component S of X. By the definition of local connectedness, S is open in X. But what about the complement of S in X? Since X is locally path connected, there is at least one open set U for every point in the complement of S in X which is connected. So now consider the union of U and S in X.)

Notice the converse is NOT true, although it's easy to erroneously assume it is. The comb space is path connected but NOT locally path connected: http://en.wikipedia.org/wiki/Comb_space

Mathemagician1234
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