This is Exercise 22.12 (b) on page 185 of Elementary Analysis, second edition, written by Kenneth Ross. I searched the site for similar questions and found a couple that looked similar (this and particularly this), but neither of those questions match mine in scope, and the language and notation used are too advanced. I am about halfway through a book called "Elementary Analysis" and I think that is a reasonable description of my mathematical skill set in regards to this question. (In other words, I don't even understand real analysis very well, so I know essentially nothing about topology.)
Background
Let $S$ be a subset of $\mathbb{R}$. Let $C(S)$ be the set of all bounded continuous real-valued functions on $S$. For $f, g \in C(S)$, let $d(f, g) = \sup \{ \lvert f(x) – g(x) \rvert \colon x \in S \}$. This makes $\big( C(S), d \big)$ a metric space.
Now I quote from Exercise 22.12 (b):
22.12 Consider a subset $\mathcal{E}$ of $C(S), S \subseteq \mathbb{R}$. For this exercise, we say a function $f_0$ in $\mathcal{E}$ is interior to $\mathcal{E}$ if there exists a finite subset $F$ of $S$ and an $\epsilon > 0$ such that
$\{ f \in C(S) \colon \lvert f(x) – f_0 (x) \rvert < \epsilon \text{ for } x \in F \} \subseteq \mathcal{E}$.
The set $\mathcal{E}$ is open if every function in $\mathcal{E}$ is interior to $\mathcal{E}$.
(b) Show the family of open sets defined above forms a topology for $C(S)$.
From Discussion 13.7 on page 87, I gather that I need to show
- $C(S)$ is open in $C(S)$.
- The empty set $\emptyset$ is open in $C(S)$.
- The union of any collection of open sets is open.
- The intersection of finitely many open sets is again an open set.
My attempt
-
Consider $f_0 \in C(S)$. We want to show that there exists a finite subset $F$ and $\epsilon > 0$ such that
$\{ f \in C(S) \colon \lvert f(x) – f_0(x) \rvert < \epsilon \text{ for } x \in F \} \subseteq C(S)$.
It seems to me that any $F$ and $\epsilon$ will suffice here.
-
Consider $f_0 \in \emptyset$. (Bear with me.) We want to show that there exists a finite subset $F$ and $\epsilon > 0$ such that
$\{ f \in C(S) \colon \lvert f(x) – f_0(x) \rvert < \epsilon \text{ for } x \in F \} \subseteq \emptyset$.
Question This is nonsense as stated. How do I conclude that $\emptyset$ is open? I would say: there are no functions in $\emptyset$. So the set of functions in $C(S)$ that are within $\epsilon$ of some non-existent function on some finite set is empty. Therefore $\emptyset \subseteq \emptyset$. Is that right?
-
Consider open sets $\mathcal{E}_1, \mathcal{E}_2, \mathcal{E}_3, \dots$ and $f_0 \in \bigcup_{i = 1}^{\infty} \mathcal{E}_i$. (We can presumably recover the finite case by letting $\mathcal{E}_n = \emptyset$ for all $n$ greater than some $N$.) We want to show that there exists a finite subset $F$ and $\epsilon > 0$ such that
$\{ f \in C(S) \colon \lvert f(x) – f_0(x) \rvert < \epsilon \text{ for } x \in F \} \subseteq \bigcup_{i = 1}^{\infty} \mathcal{E}_i$.
Consider any one $\mathcal{E}_i$ such that $f_0 \in \mathcal{E}_i$. (We know there is at least one such $\mathcal{E}_i$.) This $\mathcal{E}_i$ has a corresponding $F$ and $\epsilon$, call them $F_i$ and $\epsilon_i$. We can use $F_i$ and $\epsilon_i$ to show that the set above in curly brackets is a subset of $\mathcal{E}_i$, which means it's in the infinite union.
-
Consider open sets $\mathcal{E}_1, \dots, \mathcal{E}_n$ and $f_0 \in \bigcap_{i = 1}^n \mathcal{E}_i$. We want to show that there exists a finite subset $F$ and $\epsilon > 0$ such that
$\{ f \in C(S) \colon \lvert f(x) – f_0(x) \rvert < \epsilon \text{ for } x \in F \} \subseteq \bigcap_{i = 1}^n \mathcal{E}_i$.
There exist $F_1, \dots, F_n$ and $\epsilon_1, \dots, \epsilon_n$ such that
$\{ f \in C(S) \colon \lvert f(x) – f_0(x) \rvert < \epsilon_1 \text{ for } x \in F_1 \} \subseteq \mathcal{E}_1$.
$\vdots$
$\{ f \in C(S) \colon \lvert f(x) – f_0(x) \rvert < \epsilon_n \text{ for } x \in F_n \} \subseteq \mathcal{E}_n$.
It seems intuitively clear that $\epsilon = \min \{\epsilon_1, \dots, \epsilon_n \}$ will work. Question What do I do for $F$? I thought about using $F = \bigcap_{i = 1}^n F_i$ but I have a feeling that is wrong because I have no assurance the intersection contains anything. I thought about using $F = \bigcup_{i = 1}^n F_i$ but how do I know that any function $f$ will actually be within $\min \{\epsilon_1, \dots, \epsilon_n \}$ of $f_0$ over all elements of $\bigcup_{i = 1}^n F_i$? I guess worst case, only $f_0$ meets that criterion? In that case we could say $\{ f_0 \} \subseteq \bigcap_{i = 1}^n \mathcal{E}_i$.
Final comment
Sorry for such a long post. I am trying to be thorough and provide relevant background. I have no intuition for what I am doing in this exercise—I am just manipulating symbols. Is there any intuition for this topic that can be provided to someone at my mathematical level?
Best Answer
Your $(1)$ is fine.
$(2)$ is vacuously true: there is no $f\in\varnothing$, so there is nothing to be verified.
In $(3)$ you cannot assume that the collection of open sets is countable: the union of any collection of open sets must be open. Let $\Bbb E$ be a family of open sets, and let $\mathscr{U}=\bigcup\Bbb E$; we want to show that $\mathscr{U}$ is open. To that end let $f\in\mathscr{U}$; then there is an $\mathscr{E}\in\Bbb E$ such that $f\in\mathscr{E}$, which by definition means that there are a finite $F\subseteq S$ and $\epsilon>0$ such that
$$\{g\in C(S):|g(x)-f(x)|<\epsilon\text{ for each }x\in F\}\subseteq\mathscr{E}\;.$$
But $\mathscr{E}\subseteq\mathscr{U}$, so $\{g\in C(S):|g(x)-f(x)|<\epsilon\text{ for each }x\in F\}\subseteq\mathscr{U}$, and therefore $\mathscr{U}$ is open.
In $(4)$ your choice of $\epsilon$ is fine, but you want to take $F=\bigcup_{i=1}^nF_i$, the union of the finite sets $F_i$. Then for any $f\in C(S)$ and any $i\in\{1,\ldots,n\}$ you have
$$\begin{align*} &\{f\in C(S):|f(x)-f_0(x)|<\epsilon\text{ for each }x\in F\}\tag{1}\\ \subseteq\;&\{f\in C(S):|f(x)-f_0(x)|<\epsilon_i\text{ for each }x\in F_i\}\tag{2}\\ \subseteq\;&\mathscr{E}_i\;, \end{align*}$$
which is exactly what you want. The point is that by making $\epsilon$ at least as small as each $\epsilon_i$ and making $F$ at least as big as each $F_i$, you are ensuring that the restrictions defining the set in $(1)$ are at least as strong as those in $(2)$ that are needed to ensure that all of the functions in the set that you’re defining are also in the set $\mathscr{E}_i$.