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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?

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  • $\begingroup$ Do you mean statement (1) to be just "A topological space $X$ has a basis of path-connected sets"? $\endgroup$
    – Eric Wofsey
    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). $\endgroup$
    – Eric Wofsey
    Aug 31, 2016 at 7:45
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    $\begingroup$ 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... $\endgroup$
    – Eric Wofsey
    Aug 31, 2016 at 7:58
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    $\begingroup$ @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

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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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