# Prove $B(x,\delta)$ is open. [duplicate]

Prove $B(x,\delta)$ is open.

What this question is asking me to prove...? I don't understand nor have a clue to approach the question...

## marked as duplicate by Stefan Hamcke, Michael Albanese, Ayman Hourieh, Cameron Buie, Pedro Tamaroff♦Aug 22 '13 at 0:55

• Hint: what is the definition of an open set? – Seub Aug 22 '13 at 0:36
• Remember that $B(x,\delta)$ is a set itself. You want to show that something is a subset of another set. – Hawk Aug 22 '13 at 0:39
• Here you asked about the Hausdorff dimension.. I am confused. – Pedro Tamaroff Aug 22 '13 at 0:45
• Don't let your math.SE account to your younger sister... – Gaston Burrull Aug 22 '13 at 1:13

An open set $U$ is one such that for each point $p\in U$ there is a ball $B(p,\epsilon)\subset U$.
You're asked to prove that $B(x,\delta)$ is an open set using the definition above.
To prove that $B(x,\delta)$ you should show that for every $y\in B(x,\delta)$ we find $r>0$ such that $$B(y,r)\subset B(x,\delta)$$
Hint Draw a picture to see how to find $r$.
Let $p\in B(x,\delta)$ so $d(p,x)<\delta$ in which $d$ stands for our meter. If we set $$\epsilon=\delta-d(x,p)$$ which is positive so you can easily show that the open ball, say $B'(p,\epsilon)$ is a subset of $B(x,\delta)$. This satisfies the definition of a open set.