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I have been having a lot of trouble understanding closed sets in the context of metric spaces. I understand that this means that the set (let's say X) contains all of its limit points. But I am having tons of problems in actually solving questions involving closed sets. Here is one in particular:

Let $\{[a_n, b_n]\}_{n = 1}^\infty$ be a sequence of closed intervals such that $|a_n|\leq 1$ and $|b_n|\leq 1$ for every positive integer $n$. Prove that $\{\{x_n\}_{n= 0}^\infty\mid x_n\in [a_n, b_n]\}$ is a closed subset of $H^\infty$.

Now obviously the sequence in question will be between -1 and 1 (inclusive) and is therefore closed. But how do I state this more elegantly and mathematically?

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    What do you mean by $H^\infty$? – 2017-02-20
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    @copper.hat It denotes the set of all real sequences $\{a_n\}$ such that $|a_n| <= 1$ for every positive integer n. – 2017-02-20
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    What is the metric? Can you show that $C_n = \{ x \in H^\infty | x_n \in [a_n,b_n] \}$ is closed and then conclude that $\cap_n C_n$ is closed? – 2017-02-20
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    @copper.hat I'm sorry, this whole concept has confused me. I understood metric spaces and the distance function, but upon getting to open and closed sets I'm lost. Do i show $C_n$ is closed by proving it contains all its limit points? How do I do that? – 2017-02-20
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    What is the distance function for $H^\infty$? Yes, showing that it contains all its limit points is certainly one way, There may be easier ways, but I don't know what distance you use on $H^\infty$ so cannot offer any direct assistance. – 2017-02-20
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    A set $U$ in a metric space is open **iff** for any $x \in U$ there is some $\epsilon>0$ such that $B(x,\epsilon) \subset U$. Since $B(x,r) = \{ y | d(x,y) < r \}$, we need the distance function to figure out if a set is open. A set is closed **iff** its complement is open. Your intuition from $\mathbb{R}^n$ will carry through to a limited extent. – 2017-02-20
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    The only thing I can see is that in a prior question in the book, when we first used $H^{\infty}$ , we had to prove that $$d(\{a_n\}, \{b_n\}) = \sum_{n=1}^{\infty} \frac{|a_n - b_n|}{2^n}$$ – 2017-02-20
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    Presumably you mean $d$ is defined by this expression? This is indeed a distance. This is what is used to define the open and closed sets. – 2017-02-20
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    So would I be off base in saying that the numerator is less than or equal to 2, and thus the entire distance function is less than or equal to 1. So making epsilon equal to 1? – 2017-02-20
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    I'm not sure what you are getting at... I have added an answer below. – 2017-02-20

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One way to show that the set $C$ is closed is to show that if $x_n \to x$ with $x_n \in C$, then $x \in C$.

Note that by $x_n \to x$, I mean $d(x_n,x) \to 0$ where $d$ is given by the formula in the comments above.

Be careful with the indices here. Each $x_n$ is an element of $H^\infty$ and is itself a sequence $x_n(1),x_n(2),...$.

Let $C = \{ x | x(k) \in [a_n,b_n] \}$, $x_n \to x$ and suppose $x_n \in C$ for all $n$.

We want to show that $x \in C$.

Since ${1 \over 2^k} |x_n(k)-x(k)| \le d(x_n,x)$, and $x_n \to x$, we see that $x_n(k) \to x(k)$ for each $k$. Since $x_n(k) \in [a_k,b_k]$ for all $n$, we see that $x(k) \in [a_n,b_n]$. Since this is true for all $k$ we have $x \in C$.

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    Excellent. I think I have finally gotten closer to understanding sets. Would you say that problems like these (at least when first being taught) are usually solved using the distance function? – 2017-02-20
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    Yes. The distance function is the essence of a metric space, so it is inextricably involved. – 2017-02-20
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    thank you very much, I had been banging my head against the wall on this one for awhile – 2017-02-20
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    Delighted to assist. – 2017-02-20