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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$\begingroup$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 $A$.
$\endgroup$ 1 $\begingroup$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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