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Umbilic points on a connected smooth surface problem

Here we have a proof that if every point in a surface $S\subset\mathbb{R}^3$ is umbilical then it is contained in a sphere or a plane. But this proof only works for open sets of S.

In Manfredo's Differential Geometry of Curves and Surfaces is a proof that if this is true in a neighborhood of $p$, for all $p\in S$, then the surface is contained in a plane or sphere. It is the second part of the proof.

But at certain point he says "Since S is connected, given any other point $r$ in S there is a continuous curve in S, $\alpha: I\rightarrow S$ such that $\alpha (0)=p$, $\alpha (1)=r$.

how can this be true in general? Connectedness does not imply path connected in general, right?

Community
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Marra
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1 Answers1

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A topological space $X$ is called locally path connected if every $x \in X$ is contained in a neighborhood $U$ which is path connected.

If $X$ is locally path connected, then it is connected iff it is path connected. We know that path connected implies connected always. To see the reverse implication in this case: Suppose $X$ was not path connected. Then we can write $X$ as a disjoint union of (at least two) path-connected components. But each path component must be open, by the locally path connected criteria.

Now, any manifold is locally homeomorphic to $\mathbb{R}^n$, and since $\mathbb{R}^n$ is locally path connected, we have that any manifold is locally path connected.

And so for a manifold, connectedness does imply path-connectedness.

Alex Zorn
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