# Showing an open ball is open

Let X be a metric space with metric d. I'm trying to show that U(xo;e) is an open set. What I note so far is to talk about a subset U that for each xo an element of U, there is a corresponding e > 0 s.t. U(xo;e) is contained in U.

Does this finish the proof?

-
What does this have to do with "R^n analysis"? And what is your exact definition of U(x0;e)? –  Arturo Magidin Nov 6 '11 at 5:24
What you write does not make sense. You are trying to prove a statement about x0, but your argument uses x0 as a free variable (when you say "for each x0 an element of U"). –  Arturo Magidin Nov 6 '11 at 5:25

I'm assuming that $U(x_0,e)$ means: $$U(x_0,e) = \{ x\in X\mid d(x,x_0)\lt e\},$$ that is, the open ball with center in $x_0$ and radius $e$.

I'm also assuming that your definition of open set for a metric space is:

A set $A$ is open if and only if for every $a\in A$ there exists $\varepsilon\gt 0$ such that $U(a,\varepsilon)\subseteq A$.

If this is the case, then what you write does not even begin the proof that for a given $x_0$ and a given $e\gt 0$, $U(x_0,e)$ is open; it only repeats the definition of "open set" in a way that is confusing (using $x_0$ and $e$ for arbitrary points and radii, even though those names are already in use for a specific point and a specific radius).

Instead, what you need to do is the following: to show that $U(x_0,e)$ is open, let $y\in U(x_0,e)$ be a point in the set. What you know is that $d(x_0,y)\lt e$. What you want to show is that there exists a $\varepsilon\gt 0$ such that $U(y,\varepsilon)\subseteq U(x_0,e)$.

What does that mean? A point $z$ is in $U(y,\varepsilon)$ if and only if $d(y,z)\lt \varepsilon$. A point $z$ is in $U(x_0,e)$ if and only if $d(x_0,z)\lt e$. So you want to show that there exists $\varepsilon\gt 0$ with the property that if $d(y,z)\lt \varepsilon$, then $d(x_0,z)\lt e$.

My two word hint for that is: "Triangle Inequality."

-
(+1). Another hint would be: Draw a picture. –  Ragib Zaman Nov 6 '11 at 6:08
So, what we have then is d(xo,y) < epsilon and d(y,z) < epsilon then d(xo,z) < epsilon and our proof is complete by the Triangle Inequality? –  Buddy Holly Nov 7 '11 at 5:31
@Buddy: No, we don't have that. Notice that I used two different letters, representing two different quantities. Did you actually try to write it out? If you do, you'll see that your statement does not make sense. –  Arturo Magidin Nov 7 '11 at 5:38
I don't get how the triangle inequality fits, can you please illustrate? –  Buddy Holly Nov 7 '11 at 5:45
@BuddyHolly: When I give the hint of "Triangle inequality", this does not mean that saying "by the Triangle Inequality..." will suddenly, and magically, create a complete argument. You need to think about what the triangle inequality tells you, what you have, and what you are trying to do. Here, you are trying to say something about the distance from $x_0$ to $z$; you know something about the distance from $x_0$ to $y$, and you can control the distance from $y$ to $z$. How does the triangle inequality fit? Well, what does it tell you when you have $d(x_0,y)$, $d(x_0,z)$, and $d(y,z)$? –  Arturo Magidin Nov 7 '11 at 19:15