Every epsiolon neighborhood is an open set

[bobsagat57]

Homework Statement

In any metric space, every ε-neighborhood N(P,ε) is an open set

Homework Equations

The Attempt at a Solution

I'm completely lost on how to start this proof. I considered assuming that there existed a neighborhood that was closed and show by contradiction but from there I was uncertain how to even structure the proof. Any help would be greatly appreciated.

 

[Deveno]

first of all, what is your definition of open set? does N(P,ε) qualify?

(it's probably something along the lines of "a set is open if it contains a neighborhood of each of its points" but you may have some other definition).

 

[bobsagat57]

An set is open if every point in the set is an interior point.

P in T in an interior point if there exists some ε greater than 0 such that N(P,ε) is an element of T

And yes N(P,ε) is an open set

 

[Deveno]

don't you mean: P in T is an interior point of T if there is SOME ε-neighborhood N(ε,P) that is a SUBSET of T?

it's not kosher to just use N(ε,P) as the ε-neighborhood of an arbitrary point in N(ε,P) (besides the fact that this is circular reasoning).

let me give an example:

suppose we have the metric space R², with the metric:

d(x,y) = ||y-x||.

one possible ε-neighborhood is D = {(x,y): x² + y² < 1},

which is N(1,(0,0)).

well, the point (1/2,1/2) is clearly in D, since (1/2)²+(1/2)² = 1/4 + 1/4 = 1/2 < 1.

but N(1,(0,0)) isn't centered at (1/2,1/2) it's centered at (0,0). and N(1,(1/2,1/2)) won't work, either, as this isn't entirely contained in D.

 

[bobsagat57]

Sorry, yes I meant for it to say a subset of T

And if it helps our definition of an epsilon-neighborhood is:

An ε-neighborhood of a point P is N(P,ε) = {Q$\in$S : d(P,Q) less than ε

 

[HallsofIvy]

Suppose point q is in $N(p,\epsilon)$. Then $d(q,p)<\epsilon$. Let r= \epsilon- d(q,p)[/itex]. What can you say about N(q,r)?

 

[Deveno]

to amplify, a metric satisfies the triangle inequality. use this fact.

 

[bobsagat57]

Suppose point q is in $N(p,\epsilon)$. Then $d(q,p)<\epsilon$. Let r= \epsilon- d(q,p)[/itex]. What can you say about N(q,r)?

to amplify, a metric satisfies the triangle inequality. use this fact.

I guess I'm still not understanding what I'm supposed to do, I'm sorry.

 

[Deveno]

the triangle inequality states that for all, x,y,z in the metric space:

d(x,z) ≤ d(x,y) + d(y,z)

suppose x is our point P, and y is our point Q.

N(ε,P) is the set of all points z such that d(P,z) < ε.

HallsofIvy's post says*: suppose d(P,Q) = r.

we know that if we choose z so that d(Q,z) < ε - r, that

d(P,z) ______ by the ______ _______ ,

so z is in N(P,ε), so N(Q,___) is contained in..?

*paraphrasing, slightly different variables used.

 

[bobsagat57]

the triangle inequality states that for all, x,y,z in the metric space:

d(x,z) ≤ d(x,y) + d(y,z)

suppose x is our point P, and y is our point Q.

N(ε,P) is the set of all points z such that d(P,z) < ε.

HallsofIvy's post says*: suppose d(P,Q) = r.

we know that if we choose z so that d(Q,z) < ε - r, that

d(P,z) ______ by the ______ _______ ,

so z is in N(ε,P), so N(___,Q) is contained in..?

*paraphrasing, slightly different variables used.

So, d(P,z) < ε by the triangle inequality, so z is in N(P,ε), so N(Q,ε) is contained in the metric space as well?

Not sure if that is correct, and I'm not sure if I'm seeing how this will help to prove the original problem.

 

[Deveno]

So, d(P,z) < ε by the triangle inequality, so z is in N(P,ε)

this part is correct.

, so N(Q,ε) is contained in the metric space as well?

this makes no sense. in fact, N(Q,ε) usually isn't contained in N(P,ε) (re-read my R² example).

Not sure if that is correct, and I'm not sure if I'm seeing how this will help to prove the original problem.

try drawing a picture. draw a circle centered at P, and draw another point Q. can you draw a circle around Q that is inside your first circle? how does this relate to the general case?

you won't want to use the same epsilon. but you don't have to. you just need to find SOME other epsilon that will work. smaller is better.

 

[bobsagat57]

Maybe I understand now:

So, given Q$\in$N(P,ε), let d(P,Q)< r.
Choose z s.t. d(Q,z) < ε - r.
So, d(P,z) ≤ d(P,Q) + d(Q,z) < r + (ε - r) = ε.
Therefore N(Q,ε-r) $\subset$ N(P,ε) and every ε-neighborhood in a metric space is an open set.

Hopefully this is somewhat the right idea.

 

[Deveno]

Maybe I understand now:

So, given Q$\in$N(P,ε), let d(P,Q)< r.

d(P,Q) is a fixed real number (the "distance" between P and Q). d(P,Q) < r is the wrong way to express this.

Choose z s.t. d(Q,z) < ε - r.

why is the set of all such z a neighborhood N(Q,ε') for some ε'? this is not a hard question, but you should be sure you understand this.

So, d(P,z) ≤ d(P,Q) + d(Q,z) < r + (ε - r) = ε.
Therefore N(Q,ε-r) $\subset$ N(P,ε) and every ε-neighborhood in a metric space is an open set.

Hopefully this is somewhat the right idea.

see my comments, earlier. you're close.