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It's well known that, in $\mathbb{R}^n$:

(1) Open and Connected $\Rightarrow$ Path-connected

The proof essentially goes through the fact that (2) Every path-connected component will be open. Using this fact, we arrive at a contradiction if we suppose there are more than one path-connected component.

Well, trying to see where (1) would keep being valid, I arrived at the following:

If $X$ is a locally convex topological vector space, then (1) is valid.

But, to prove this, I proved (2). But for that, you use the (rather strong imo) convexity of a local base (to essentially repeat the argument for $\mathbb{R}^n$). But I'm not satisfied with this, as I think that we are using a lot of strong conditions (using the existence of a "line segment" to prove the existence of a "curve" seems very bazooka-like to me).

Anyway, given the previous considerations, my questions are:

Are there more general examples of spaces where (1) holds? Is there a characterization of spaces that satisfy this property?

And bonus question, since (2) implies (1) if the set is open.:

Are there more general examples of spaces where open sets satisfy (2)? Is there a characterization spaces that satisfy this property?

Aloizio Macedo
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    In a locally path-connected space, connected open sets are path-connected. Every Topological vector space over $\mathbb{R}$ or $\mathbb{C}$ is locally path-connected, since the balanced neighbourhoods form a local base at $0$. – Daniel Fischer Apr 23 '14 at 19:56
  • Thank you, that takes care of half the trouble. What about topological spaces in general? – Aloizio Macedo Apr 23 '14 at 20:02
  • Nothing other than local path-connectedness springs to mind. – Daniel Fischer Apr 23 '14 at 20:17
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    Note that local path connectedness is not necessary. Consider the rationals, which are not locally path connected, but every connected open set is path connected (because every such set is empty). Your property 2 also fails here: the path components are singletons and are not open. – Nate Eldredge Apr 23 '14 at 20:25

2 Answers2

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(this is still a partial answer)

The following two properties of a topological space $X$ are equivalent (I use your numeration):

(2) every path-connected component of an open subset of $X$ is open

(3) $X$ is locally path-connected

Proof. (3) implies (2). Take a point $x$ of an open subset $U$, and consider the path-connected component $Y$ of the subset that contains $x$; since the space is locally path-connected, there exists a path-connected open neighbourhood $V$ of $x$ contained in $U$; obviously $V$ must be contained in $Y$, so $Y$ is open.

(2) implies (3). Take a point $x$ and an open neighbourhood $U$ of it. By hypothesis the path-connected component of $U$ containing $x$ is open so we are done.

The following property is not equivalent to (2)-(3), but is implied by them:

(1) open connected subsets of $X$ are path-connected.

As Nate suggested, the set of the rational number with the euclidean topology trivially satisfies (1) but doesn't satisfy (2)-(3).

Davide Cesare Veniani
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I think that a topological space X is path connected if and only if the following two properties are true: (a) X is connected (b) every point in X has a path connected neighborhood.

On the other hand, X is locally path connected if and only if any open subset of X has property (b).

Bill Cano
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  • In response to the "I think...": Yes, Bill's statement here is correct. See (https://www.nsm.buffalo.edu/~badzioch/MTH427/_static/mth427_notes_8.pdf) for converse direction, and forward direction is a simple exercise to show. – J.G.131 Aug 10 '24 at 19:27