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While reading about general topology, I started looking at the metric topology. At one point, the interior of a topological space $X$ is defined as such:

$${E}^{\circ}=\{p\in E\ | \ B_p(\epsilon)\subseteq E, \epsilon>0\}$$

Here, $B_p(\epsilon)$ is a ball with center $p$ and radius $\epsilon$. I tried verifying that this met the general properties of an interior, and didn't have much trouble showing ${E}^{\circ}\subseteq E$, $(E\cap F)^{\circ}={E}^{\circ}\cap {F}^{\circ}$, and ${X}^{\circ}=X$. However, I don't see how ${{E}^{\circ}}^{\circ}={E}^{\circ}$. I see that ${{E}^{\circ}}^{\circ}\subseteq {E}^{\circ}$, but tried without success to see the other containment. Is there a reason why this follows simply from the axioms of what a metric is?

Futhermore, apparently another way to approach this topology is to define the family of open sets $\mathcal{T}$ as those which are the union of a family of balls. Again, I tried verifying this for myself. I see that for any $\mathcal{U}\subseteq\mathcal{T}$, $\cup\mathcal{U}$ is a union of sets which each may be represented as the union of a family of balls, so the whole union again may be represented as the union of a family of balls, namely those in each set in $\mathcal{U}$. Also, $X$ may be represented as the union of balls where I take one ball with center $p$ for all $p\in X$. Finally $\emptyset$ is just an empty union of balls. However, I was unable to show that for finite $\mathcal{U}\subseteq\mathcal{T}$, $\cap\mathcal{U}$ is again open. Is there some way to see that the finite intersection could be represented as a union of balls from the properties of the metric?

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up vote 3 down vote accepted

A hint for your first question: show that if $p \in E^{\circ}$, then there exists an $\epsilon > 0$ such that $B_p(\epsilon)$ is contained in $E^{\circ}$. This shows that $p$ lies in the interior of $E^{\circ}$, i.e., $p \in (E^{\circ})^{\circ}$. This comes down to showing that if you have a point $p$ in an open ball $B$, you can find a smaller opne ball $B'$ around $p$ which is entirely contained in $B$.

A hint for your second question: it's enough to show that if $B_1$ and $B_2$ are two open balls and $p \in B_1 \cap B_2$ (i.e., $p$ lies in both balls), then there is a small ball $B_3$ around $p$ such that $B_3$ is contained in both $B_1$ and $B_2$. This is actually quite similar to the exercise of the previous paragraph.

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I know it's been a while since I posted this, but I recently found the time to take your hints (very clear to me btw) to heart and figure this out. I know it's a bit overdue, but thank you. – yunone Nov 15 '10 at 9:08

For the first problem, show that for every point in $E^{\circ}$ there is a ball around it which is also in $E^{\circ}$. In other words, show that interiors are open.

For the second problem, do the same: show that for every point in the intersection there is a ball around it which is also in the intersection.

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Sound advice. :) – Pete L. Clark Nov 3 '10 at 8:31

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