# Infimum of a metric space

1. Feb 8, 2006

### mathboy20

Hi

I have another question in the field of analysis.

$$Y \subseteq \mathbb{R}^n$$

I'm suppose to show that if $$x \in \mathbb{R}^n$$, then the set

$$\{ || x - y || \ y \in Y \}$$

has an infimum, such that

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

I know that I'm suppose to show that the infimum is the shortest distance between x and y. But how I proceed from there?

Where $$f: \mathbb{R}^n \rightarrow \mathbb{R}$$.

Sincerely Yours
Mathboy20

Last edited: Feb 8, 2006
2. Feb 8, 2006

### matt grime

Lemma: any bounded below set of real numbers has an inf.

Proof: if not I can create a sequence in the set tending to minus infinity.

3. Feb 8, 2006

### mathboy20

If I have understood You correctly, I then need to show, that its impossible to create a sequence which tends to minus infinity?

Sincerely Yours

Mathboy20

4. Feb 8, 2006

### matt grime

No, you have not understood me correctly.

The set of distances is a subset of [0,infinity) so it is necessarily bounded below and has an inf. All bounded below sets have infs. That is one of the elementary facts you learn about infs. (elementary in the sense of 'the first things', not necessarily the easiest.)

You do not need to reprove this fact, I was merely illustrating why the fact was true, hoping to jog your memory.

Last edited: Feb 8, 2006
5. Feb 8, 2006

### mathboy20

Okay I think I get it now :)

So what I need to show is that the distance set {||x-y||} is bounded below, and by showing this it implies (according to the definition) that the distance set has a greatest lower bound aka infimum?

Definition of lower bound:

A number less than or equal to any number in a set

Proof:

Since $$x \in \mathbb{R}^n$$, and $$y \in Y \subseteq \mathbb{R}^n$$

Then $$x \in \{||x-y|| \}$$

Ergo $$\{||x-y|| \}$$ has a lower and therefore according to the definition of glb also an infimum.

Sincerely Yours

Mathboy20

Last edited: Feb 8, 2006
6. Feb 8, 2006

### matt grime

Yes, and distances are trivially (ie the definition of a metric ensures that this is true) bounded below by zero. To be totally rigorous you might want to add that since Y is non-empty the set of distances is a non-empty bounded below set and so has an inf.

Last edited: Feb 8, 2006
7. Feb 8, 2006

### mathboy20

thank You very much Matt,

Sincerely Yours

Mathboy20

8. Feb 8, 2006

### mathboy20

I final question regarding this matter.

If $$x \in Y$$ then I'm suppose to show, that f(x) = 0

Any hints on how I do that?

Sincerley Yours

Mathboy20

Last edited: Feb 8, 2006