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Is a compact, connected, and "locally convex" set in $\mathbb{R}^n$ convex?

Here I mean a space $A$ locally convex as: For any point $x\in A$, there exists a neighborhood $U$ of $x$ s.t. $U$ is convex. This question comes from a conversation with my friend, and we cannot find a way to prove or disprove it since we know few techniques regarding convexity. Things we have noticed are:

  1. Compactness cannot be deleted because annulus in $\mathbb{R}^2$ without boundary points is a counterexample.

  2. Connectness cannot be deleted because separated open disks in $\mathbb{R}^2$ is a counterexample.

Besides this, we obtain few results. It would be appreciated if someone could provide related texts or techniques.

skt_zheng
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1 Answers1

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Yes, such a set is convex. To see this, note first that $A$ is locally connected via piecewise linear paths, and hence globally connected by such via connectedness (this is the same argument that upgrades local path-connectedness to global path connectedness in a connected space- namely, the PL-connected components are open and disjoint, so by connectedness there can be only one). In particular, $A$ is rectifiably connected.

There is therefore a path joining any two points with minimal length* (this follows from compactness, equicontinuity of arc length parametrizations, and lower semi continuity of length, see e.g. Semi-continuity of arc length)

But a minimal length path joining two points in $A$ must be locally linear (hence linear), elsewise it can be shortened at some point by local convexity.

*Update

To elaborate more on this argument, if $\gamma_i$ is a sequence of arc-length parametrized paths that approaches the infimum $L$ of path-lengths between $p$ and $q$, then extend each $\gamma_i$ so that they all have a common domain $[0,L']$, where $L'\geq L$ is the largest length of any path in the sequence (we extend in such a way that each path is constantly equal to the endpoint $q$ on the extension).

Now the paths form a sequence of $1$-Lipschitz maps, and are hence equicontinuous, so by the Arzelà–Ascoli theorem, there is a subsequence converging pointwise to a $1$-Lipschitz map $\gamma\colon [0,L']\to A$. Since the lengths approach $L$, from the way we defined the extension we must have $\gamma(t)=q$ for each $t\geq L$, so the restriction $\gamma|_{[0,L]}$ is a $1$-Lipschitz path from $p$ to $q$, hence has length at most $L$ (and therefore exactly $L$, since $L$ was the infimal length).

M W
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  • Could you elaborate more on this answer? Part with rectifiably connected, minimal length path, and so on. – Jakobian Sep 19 '23 at 14:35
  • Brilliant! Thanks a lot for your marvelous answer, M W. – skt_zheng Sep 20 '23 at 02:27
  • I noticed an answer on this page mentioned Tietze-Nakajima Thoerem which states that compactness can be weakened to closedness. In Euclidean space, this means boundedness is unnecessary. I checked your answer and no adjustment is needed. Here is a revisit of Tietze-Nakajima Theorem. Thanks again. – skt_zheng Sep 20 '23 at 02:49
  • @skt_zheng that makes sense, thanks for the link, we certainly get pointwise compactness in Arzela Ascoli simply from the fact that all the paths are from $p$ to $q$ and have uniformly bounded length, hence after we fix our two points, everything is happening inside of $A\cap B$ for some closed ball, so as you said its certainly enough for $A$ to be an arbitrary closed set. – M W Sep 20 '23 at 20:26