Why do we specify chart image is open? (1 Viewer)

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Hi,

When we define a manifold, and in particular define what a chart is, one of the conditions we specify is that the image [tex] \phi(U) [/tex] is open in [tex]R^n[/tex]. Why do we specify this?

For example if we didn't specify this and allowed closed balls, then we could cover [tex] S^1 [/tex] by [tex] \theta [/tex] where [tex]\theta \in [0,2\pi) [/tex], and wouldn't need two charts.

I know that open balls and continuity go hand in hand, so I understand if we take [tex] U \subset M [/tex] as an open interval and want to define a continuous map [tex] \phi [/tex] then it must be that [tex] \phi(U) \subset R^n [/tex] is open. But what if our [tex] U \subset M [/tex] is not open, then why do we care if the image is open or not?

Basically why is this part of the definition?
 

Fredrik

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I can't think of a good answer right now, so I'll just point out that [itex]f:X\rightarrow Y[/itex] is said to be continuous if [itex]f^{-1}(V)[/itex] is open for each open [itex]V\subset Y[/itex], and said to be open if [itex]f(U)[/itex] is open for each open [itex]U\subset X[/itex].
 
411
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I can't think of a good answer right now, so I'll just point out that [itex]f:X\rightarrow Y[/itex] is said to be continuous if [itex]f^{-1}(V)[/itex] is open for each open [itex]V\subset Y[/itex], and said to be open if [itex]f(U)[/itex] is open for each open [itex]U\subset X[/itex].

Yep, sorry was being a bit sloppy above.
 

DrGreg

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For example if we didn't specify this and allowed closed balls, then we could cover [itex] S^1 [/itex] by [itex] \theta [/itex] where [itex]\theta \in [0,2\pi) [/itex], and wouldn't need two charts.
But then you wouldn't necessarily have the required continuity at [itex]\theta = 0 [/itex]
 
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bcrowell

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I think DrGreg's #4 works.

As an alternative way of looking at it, you want to be able to define a metric at a given point P in a chart. Say you want to express the metric as a line element [itex]d\ell^2=g_{ab}dx^a dx^b[/itex]. You need to be able to fit those infinitesimally small dx's inside a neighborhood of P, without leaving the domain of your coordinate chart. If the chart is defined on an open set, then you're guaranteed that they fit. If it's not, then P could be a boundary point, and you would then have at least one dxa that didn't fit inside the chart.

Even if you're not interested in defining a metric -- say you just want to talk about manifolds as abstract topological spaces. The standard definition of a manifold is something that "looks like" (i.e., is homeomorphic to) [itex]\mathbb{R}^n[/itex] locally. If you don't interpret "locally" to mean "on some sufficiently small *open* set," then you don't correctly capture the intended definition of a manifold, as opposed to a manifold with boundary or something else.
 

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