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So I am working my way through the Dover book, Intro to Topology by Bert Mendelson, and in the section on open and closed sets, I'm stuck on the following notation for this problem:

Let $(X,d_1)$, $(Y,d_2)$ be metric spaces. Let $f\colon X\to Y$ be continuous. Define a distance function $d$ on $X\times Y$ in the standard manner. Prove that the graph $\Gamma_f$ of $f$ is a closed subset of $(X\times Y,d)$.

(The "graph" of $f$ is the set $\{(x,y) \in X \times Y : y = f(x)\}$).

What I don't understand what it means to prove that the graph of f is a "subset" of $(X \times Y, d)$. Obviously the graph is a subset of $X \times Y$. From what I understand it means, "the set $X \times Y$ with a metric $d$." How does the metric part play into to proving that it is a closed subset?

On a side note, I thought all I should prove is that the graph of $f$ is a subset of $X \times Y$ and that subset is closed, but I'm not so sure. Thanks for the help!

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It really depends on the book how "closed" gets defined. The definitions are all equivalent, however. If the book has not defined closed for metric spaces, has it defined "open?" – Thomas Andrews Nov 22 '12 at 3:37
The book has defined what a closed set is, however I am more concerned specifically with the metric part. How does that play into the proof? – Eric Nov 22 '12 at 3:49
Yes, it’s clear that $\Gamma_f\subseteq X\times Y$; you’re being asked to prove that this subset is closed in $X\times Y$ in the topology on $X\times Y$ generated by the metric $d$. As Thomas said, exactly how you’re expected to do this depends on exactly how you’ve defined closed. It may, for instance, be enough to show that every sequence in $\Gamma_f$ that converges in $X\times Y$ has its limit in $\Gamma_f$. Or you may have to show that if $p=\langle x,y\rangle\in(X\times Y)\setminus\Gamma_f$, then there is an open ball $B_d(p,\epsilon)$ about $p$ disjoint from $\Gamma_f$. – Brian M. Scott Nov 22 '12 at 3:49
The metric $d$ defines the topology on $X\times Y$, which in turn determines which sets are closed. – Brian M. Scott Nov 22 '12 at 3:51
Some books define closed directly in terms or the metric, others define "open" in terms of the metric, and then defined "closed" in terms of "open." – Thomas Andrews Nov 22 '12 at 3:54
up vote 2 down vote accepted

A subset $A$ of a set $X$ is closed with respect to a metric $d$ on $X$ precisely if every point in the closure of $A$ is in $A$. A point $a\in X$ is in the closure of $A$ if for every positive number $\varepsilon$, no matter how small, the ball of radius $\varepsilon$ centered at $a$ intersects $A$.

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Wow, I got a lot of great answers! If I could accept everyone's response as an answer I would Thanks everyone! – Eric Nov 22 '12 at 4:05

An alternative but very important way to define a closed subset of a metric space X is as the complement of an open subset of X. The definition of an open subset is defined here:

You can then define a closed subset A of X as any set where it's complement X\A is open in X.

This definition of closed subset of X is equivalent to Michael's definition (and as an exercise you should try to prove that the two definitions are equivalent).

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