Troubleshooting Metric Space Problems: Infimum and Closed Sets

[Mathman23]

Hi

I have this here metric space problem which caused me some trouble:

$$S \subseteq \mathbb{R}^n$$ then the set

$$\{ \| x - y \| \ | y \in S \}$$ has the infimum $$f(x) = \{ \| x - y \| \ | y \in S \}$$

where f is defined $$f: \mathbb{R}^n \rightarrow \mathbb{R}$$
I have two problems here which I'm unable to solve:

(a) show, if S is a closed set and $$x \notin S$$ then $$f(x) > 0$$ ?

(b) show, if S is a closed set, then $$S = \{ x \in \mathbb{R}^n | f(x) = 0\}$$ ?

I need to hand this in tomorrow, and I have been strugling this these two problems the last week, therefore I would very much appreciate if anybody could give me an idear on how to solve the two problems above.

God bless,

Best Regards,
Fred

 

[quasar987]

$$f(x) = \{ \| x - y \| \ | y \in S \}$$ ??

That would mean to each x in R^n, f maps x to ||x-y|| for all y in S. So as soon as card(S)>1, f is not a function.

Also, what do you mean by "$$\{ \| x - y \| \ | y \in S \}$$ has the infimum f(x)"?

 

[Mathman23]

$$f(x) = \{ \| x - y \| \ | y \in S \}$$ ??

That would mean to each x in R^n, f maps x to ||x-y|| for all y in S. So as soon as card(S)>1, f is not a function.

Also, what do you mean by "$$\{ \| x - y \| \ | y \in S \}$$ has the infimum f(x)"?

Sorry it should have said

$$f(x) = \mathrm{inf} \{ \| x - y \| \ | y \in S \}$$

Any idears on how to go about this?

Best Regards

Fred

p.s. My problems deals with the distance from $$\mathbb{R}^n$$ to a point in a subset S of $$\mathbb{R}^n$$.

 

[quasar987]

a) wouldn't that exeedingly simply argument suffice:

we know that ||x-y|| = 0 iff x=y. But since x is not in S, x is not equal to y for any y in S. Hence, ||x-y||>0.

There's probably a problem with this argument as it doesn't even use the closedness of S...

 

[Mathman23]

Hello and thank You for Your answer,

Then (B) is that the oposite of (A) ??

Sincerely and God bless

Fred

a) wouldn't that exeedingly simply argument suffice:

we know that ||x-y|| = 0 iff x=y. But since x is not in S, x is not equal to y for any y in S. Hence, ||x-y||>0.

There's probably a problem with this argument as it doesn't even use the closedness of S...