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

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#### 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

Staff Emeritus
Gold Member
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

Gold Member
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 $\theta = 0$

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#### bcrowell

Staff Emeritus
Gold Member
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 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) $\mathbb{R}^n$ 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.

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