Let $F = {f_{n} | n ∈\Bbb N }$ be an infinite collection of functions $f_{n}(x)=e^{−n(x−n)^2} , x ∈ \Bbb R$. I am trying to prove that $F$ is closed in $BC(\Bbb R, \Bbb R)$, where $BC$ is the space of bounded continuous functions with $sup$ metric.
In order to do this, one way would be to estimate the distance between two functions , say $f_{m} $ and $f_{n}$ in this family by a constant say $1/2$. Doing this would imply set being discrete, thereby making all subsets open and closed.
But I am running into a problem shown below:
$$|f_{m}(x)-f_{n}(x)|=|e^{−m(x−m)^2}-e^{−n(x−n)^2}|\leq |e^{−m(x−m)^2}|+|e^{−n(x−n)^2}|=2$$
So, I am not getting an estimate by $1/2$. I am getting $2$ instead.
Any modifications suggested?
I think you've got your inequality wrong. You want to find a positive constant $\epsilon$ so that $||f_n- f_m|| \ge \epsilon$ whenever $n\neq m$. (So that $F$ is discrete and thus closed). By definition,
$$||f_n- f_m|| = \sup_{x\in \mathbb R} |f_n(x) - f_m(x)|. $$ In particular, for any $x\in \mathbb R$,
$$||f_n- f_m|| \ge |f_n(x) - f_m(x)|.$$
Now, $f_n (x) = e^{-n(x-n)^2}$ as a maximum at $x = n$. And whenever $m\neq n$,
$$f_m(n) = e^{-m(n-m)^2} \leq e^{-m} \le e^{-1}$$
(The last inequaility holds as $m \geq 1$). So
$$||f_n - f_m|| \ge |f_n(n) - f_n(m)| = 1 - e^{-m(n-m)^2} \ge 1 - e^{-1}$$
and so $F$ is discrete.