Proving Distance Between a Point and a Closed Set

[henry22]

Homework Statement

Prove that a point,a, not belonging to the closed set B has a non-zero distance from B. I.e that dist(a,B)=inf(y in a) ||a-y||>0

Homework Equations

I have no idea how to start this. It is only worth a few marks and I have been told it is fairly easy but I have always struggled with proofs.

The Attempt at a Solution

Would I use contradiction somehow?

 

[Petek]

Are you working in a metric space? Also, you might want to check your definition of dist (a, B). It's not quite right. Anyway, here's a hint: If B is closed, then its complement is ... .

 

[henry22]

Sorry yes the inf part should be (y in B) not y in a. So if B is closed the complement must be open. So the point a is in an open set?

 

[Petek]

Look in your text! You should see something like

dist(a, B) = inf (dist(a , b) : b $\in$ B)

And yes, a belongs to an open set that doesn't intersect B. Therefore, there exists an open ball centered at a such that ... .

 

[henry22]

Such that (x in X : ||x-a|| < r)?

 

[micromass]

So you've proven that there exists a ball with radius r around a, such that B doesn't intersect that ball.
Now, take a point b in B. Is b an element of our ball? What can you say now over the distance between a and b?

 

[henry22]

b does not intersect the ball and so the distance between a and b will be >= r-b ?

 

[micromass]

Where do you get r-b?? The distance between a and b is always greater than r...
So, whatever b we take, we got dist(a,b)>=r. What can you say now about dist(a,B)??

 

[henry22]

sorry I forgot we had centred the ball at a, I was thinking a was any point.

So now we can say that dist(a,B)>=r as well

 

[micromass]

Seems like you've got it then!

 

[henry22]

Is it that easy?! I was expecting more work than that to be honest ha.

thanks for the help guys