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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?

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    What you wr$i$te does not make sense. You are try$i$ng to prove a statement about $x$0, but $y$our argument uses x0 as a free variable (when you say "for each x0 an element of U").2011-11-06

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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."

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    @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 $d(y,z)$?2011-11-07