Why do we specify chart image is open?

[LAHLH]

Hi,

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

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

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

Basically why is this part of the definition?

 

[Fredrik]

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

 

[LAHLH]

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

Yep, sorry was being a bit sloppy above.

 

[DrGreg]

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

But then you wouldn't necessarily have the required continuity at \theta = 0

 

[bcrowell]

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 d\ell^2=g_{ab}dx^a dx^b. 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 dx^a 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) \mathbb{R}^n 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.