Finding closure/interior of subset of function space Consider the subset $$A=\left\{f\in C(\Bbb R): |f(x)|&lt; \frac{1}{1+|x|} \, \text{for all } x\in \Bbb R\right\}\subset \left\{f\in C(\Bbb R): \lim_{|x|\to \infty}f(x)=0 \right\}=X.$$

where $X$ is equipped with the operator norm. What is the closure and interior of $A$?


  My guess is that the closure of $A$ is the set $$B=\left\{f\in C(\Bbb R): |f(x)|\leqslant \frac{1}{1+|x|} \, \text{for all } x\in \Bbb R\right\}$$
  and the interior is the set
  $$C=\left\{f\in C(\Bbb R): \exists\epsilon&gt;0 \text{ such that }|f(x)|&lt; \frac{1}{1+|x|}-\epsilon \, \text{for all } x\in \Bbb R\right\}.$$


I'm fairly certain I got the closure right, but I'm unsure about the inerior.

Certainly $C$ is open and a subset of $A$, therefore, $C\subset \operatorname{int}A$. Does $\operatorname{int A}\subset C$ also hold?

Are my guesses correct?
According to your definition of $C$, $A\cap C =\emptyset$ ( if $f\in A$, $|f|$ eventually becomes smaller than $\epsilon$.)

Perhaps it makes more sense to consider the set:

$
C=\left\{ f\in A\,\,\, |\,\,\, \exists M&gt;0 \,\,\,s.t.\,\,\, f(x)=0\,\,\, \forall |x| \geq M \right\}.
$
According to your definition of $C$, $A\cap C =\emptyset$ ( if $f\in A$, $|f|$ eventually becomes smaller than $\epsilon$.)

Perhaps it makes more sense to consider the set:

$
C=\left\{ f\in A\,\,\, |\,\,\, \exists M&gt;0 \,\,\,s.t.\,\,\, f(x)=0\,\,\, \forall |x| \geq M \right\}.
$