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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?

Eric Wofsey
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Bob
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  • Do you mean statement (1) to be just "A topological space $X$ has a basis of path-connected sets"? – Eric Wofsey Aug 31 '16 at 07:28
  • @Eric: No, I typed what I meant but what I meant was wrong apparently. – Bob Aug 31 '16 at 07:43
  • 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). – Eric Wofsey Aug 31 '16 at 07:45
  • @Eric: Point taken, I made an edit that I hope makes my question more clear. – Bob Aug 31 '16 at 07:53
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    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... – Eric Wofsey Aug 31 '16 at 07:58
  • Since you accepted my answer, it seems my interpretation was correct. I'm going to go ahead and clarify the question. – Eric Wofsey Aug 31 '16 at 08:14
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    @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 – 57Jimmy Aug 31 '16 at 08:17

1 Answers1

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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).

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Eric Wofsey
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