I Compact manifold cannot be represented by (single) parametric equation

elias001
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Show that a compact manifold cannot be represented by a (single) parametric equation.
Show that ## M ## is a compact manifold in ##\mathbb{R}^{n},\ ## then ##\partial\ M\ ## is also compact; if also ## M ## is ## n ##-dimensional, then ##\partial\ M=## bdry ## M.##

Show that a compact manifold cannot be represented by a (single) parametric equation.

I asked online about two portion questions exercise, and I would like to know if the solutions displayed below is correct?

We know that ##M## is compact manifold in ##\mathbb{R}^n## and the boundary (of any manifold with boundary) ##\partial M## is closed in ##M##. Since every closed subset of a compact space is compact, then ##\partial M## is compact. For the problem that any compact manifold cannot be represent as single parametric equation, just note that if we can, then ##M## must be homeomorphic to an open subset of ##U \subset \mathbb{R}^{\text{dim }M}##. That is, there exists homeomorphism ##\varphi : M\to U = \varphi(M)\subset \mathbb{R}^{\text{dim }M}##. This implies that ##U## is compact (closed and bounded) and also open. Since ##U \subset \mathbb{R}^{\text{dim }M}## is open and closed and ##\mathbb{R}^{\text{dim }M}## connected, this means ##U=\mathbb{R}^{\text{dim }M}##. But this is impossible since ##U=\varphi(M)## is compact by continuity.

Thank you in advance
 
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Sorry, why would ##U## need to be open?
 
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@dextercioby I am not sure if the entire solution is correct. I am not sure if ##U## need to be open myself.
 
actually the parametric mapping t-->e^it represents the compact manifold S^1 rather well.
 
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dextercioby said:
Sorry, why would U need to be open?

All definitions of parametrization I were taught were having an open subset as a domain. I checked lecture notes of my superivsor (link to pdf, but it's in polish, so beware...) and it says on page 9 that "parametrization has to be an injection..." and something more about differentiability, but I guess this thread is about topological manifolds only, so this doesn't matter.
 
@weirdoguy, @mathwonk, @dextercioby I did more googling, and this is what I found, someone asked online "I am learning about differentiable manifold, and I got a question asking to give a reason why I need at least two charts to cover a compact manifold." His idea is:

If ##M## is a compact manifold covered by a single chart ##(U,\phi)##, then by the definition (the locally Euclidean), ##U## is homeomorphic to the open subset ##\phi(U) \subset R^{n}## for some ##n##.

But the only connected open compact subset of ##R^{n}## is empty set for ##n \geq 1##, if ##n=0## we also have ##R^{0}## itself. Then, since ##\phi(U)## is not empty, it cannot be compact.

This idea seems right in the first place, but afterwards a confusion arose. Yes, I prove that ##\phi(U)## is not compact, but it does not contradict the homeomorphism, since ##M## is compact manifold, but ##U## is not necessarily compact. Or a compact manifold is naturally covered by a compact covering? or the covering itself is naturally compact?

After much discussion with others online, he got the key insight:

Since ##U## is a chart, ##U\subset M##. Since ##U## covers ##M##, ##M\subset U##. So ##M=U##

he gave the answer as follows:

Since ##U## is a chart, then ##U \subset M##, but since ##U## covers ##M##, ##M \subset U##. Thus, ##M=U##.

Thus the chart now is actually ##(M,\phi)##.

By the definition of local Euclidean, ##M## is homeomorphic to an open subset ##\phi(M) \subset R^{n}##.

But the only connected open compact subsets of ##R^{n}## are empty set and ##R^{0}## if ##n=0##, but ##\phi(M)## is not empty, thus ##\phi(M)## is not compact, which contradicts the homeomorphism.
 
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