4
$\begingroup$

Why are the following two statements equivalent for any topological space $X$?

1) $X$ is locally path connected (meaning, it has a basis of path connected sets).

2) Every point of $X$ has a path connected neighborhood.

Is it simply that a path connected neighborhood is an open set in the subspace topology?

$\endgroup$
7
  • $\begingroup$ Do you mean statement (1) to be just "A topological space $X$ has a basis of path-connected sets"? $\endgroup$ Aug 31, 2016 at 7:28
  • $\begingroup$ @Eric: No, I typed what I meant but what I meant was wrong apparently. $\endgroup$
    – Bob
    Aug 31, 2016 at 7:43
  • $\begingroup$ No, but what you wrote really really really doesn't make sense. Statement (1) is simply a true statement: if $X$ has a basis of path connected sets, then $X$ is locally path connected. The truth of (2), on the other hand, depends on what the space $X$ is. So to say that (1) and (2) are equivalent would be to say that every topological space satisfies condition (2). $\endgroup$ Aug 31, 2016 at 7:45
  • 1
    $\begingroup$ I still highly doubt that what you've written is what you actually want to ask. Is the following what you really want to ask: "Why is it true that for any topological space $X$, $X$ has a basis of path-connected sets iff every point of $X$ has a path connected neighborhood?" That's the question I answered... $\endgroup$ Aug 31, 2016 at 7:58
  • 1
    $\begingroup$ @Bob This is an instance of the common topological problem of "local definitions"; there are some forms which are used to define local properties, and they are in general NOT equivalent: 1) every point has a * neighborhood (this always holds if the space itself is globally *) 2) every point has an open * neighborhood 3) there is a basis of * sets $\endgroup$
    – 57Jimmy
    Aug 31, 2016 at 8:17

1 Answer 1

4
$\begingroup$

They aren't equivalent. Indeed, any path-connected space satisfies (2), since you can take the neighborhood to just be $X$ itself. But not every path-connected space is locally path-connected (see https://math.stackexchange.com/a/135483/86856, for instance).

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.