# Is this set really not closed ? ?

1. Feb 7, 2006

### quasar987

Is this set really not closed ?!!?!

Ok, consider the metric space $\mathbb{R}^2$ armed with the pythagorean metric $d(x,y) = ||x-y||$ and B, the closed ball of radius 1 centered on the origin: $B = B_1(0) \cup \partial B_1(0)$. Now construct the metric space composed of this closed ball and the appropriate restriction on the pythagorean metric.

Now consider D, a closed ball centered on the origin but of radius r<1, and ask the question: is D closed in B? D is closed iff $D^c = B \slash D$ is open in B. But it is not, for consider a point on the border of B ($x_0 \in \partial B_1(0)$). No ball with x_0 as center is contained entirely in B. Hence D is not closed in B.

Is that correct?!

2. Feb 7, 2006

### AKG

No, it's not correct; D is closed in B. If B is your metric space, then the ball of radius s centered at x0 is defined as:

$$\{x\ \mathbf{\in B}\, :\, ||x - x_0|| < s\}$$

They key part is the "$\in B$" part. So a ball in B is the intersection of the ball in R² with the space B itself.

3. Feb 8, 2006

### quasar987

Thx AKG.

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4. Feb 9, 2006

### quasar987

But then how would any set NOT be open?

G is open if for all x in G, there exists r>0 s.t. B(r,x) is in G.

But the definition of B(r,x) is all g in G such that d(x,g)<r. So for any r, B contains only elements of G, so B(r,x) is necessarily in G, making G open.

what the heck?!

I would appreciate help quickly. I leave for exam in 13 minutes exactly. *sweat*

Last edited: Feb 9, 2006
5. Feb 9, 2006

### quasar987

At least there were no questions on open sets. phew.

6. Feb 9, 2006

### Hurkyl

Staff Emeritus
No: if X is your metric space, then B(r,x) is all g in X such that...

7. Feb 9, 2006

### matt grime

very simply, in a metric topology. The set of things distance less than or equal to 1 from some given point would not be open in general.

8. Feb 9, 2006

### quasar987

Oh, of course!

9. Feb 10, 2006

### HallsofIvy

Staff Emeritus
If G is the entire set, (as you say "the definition of B(r,x) is all g in G such that d(x,g)<r") then yes, it is open. The entire base set in a topological space is always both open and closed.

If you mean that G is a proper subset of some set X on which the topology is defined then "the definition of B(r,x) is all g in G such that d(x,g)<r." is not true. B(r,x) is all g in X such that d(x,g)<r.

10. Feb 10, 2006

### quasar987

Exactly.

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