Why is the set {x: d(x,p)>r} open in a metric space?

[golriz]

" Let X be a metric space and p be a point in X and be a positive real number. Use the definition of openness to show that the set U(subset of X) given by:
U = {x∈X|d(x,p)>r} is open. "

I have tried:
U is open if every point of U be an interior point of U. x is an interior point of U if there is an open ball B(x, r) is a subset of U. Let y belongs to B(x, r) then d(x, y)< r, and p doesn’t belong to B(x, r) since d(p, x) > r. If we can show that d(y, p) > r (y is in X) then since y is a arbitrary point of B(x, r) it means that the open ball is a subset of U.
But I don’t know how to show d(y, p) > r ?
Please help me.

 

[micromass]

What you need to do is take an arbitrary x in U. This x will have as property that d(x,p)>r.

Now you need to find an $\varepsilon>0$ suitably such that

$$B(x,\varepsilon)\subseteq U$$

So for each y with $d(x,y)<\varepsilon$ must hold that d(y,p)>r.

Now, draw a picture first. What must $\varepsilon$ be??

 

[golriz]

d(x, p) + d(p, y)> d(x, y)
d(x, y)< 2r < ε
ε = 2r
Is it correct?

 

[micromass]

No, and I don't see what you did actually...

 

[golriz]

I use triangle inequality, so:

d(x, p) + d(p, y)> d(x, y)

 

[micromass]

What you know is that $d(x,y)<\varepsilon$. What you must prove is that d(y,p)>r. I don't see how your argument solves that.