An open set in $l^{\infty}$ Let $X$ denote the set of all bounded real sequences, equipped with the norm $\| (x_n)\|_\infty:= \sup\{|x_1|,|x_2|,|x_3|,\ldots\}$; Let $X_{++}$ denote the set of all bounded positive real sequences (“sup” is supremum).
Is $X_{++}$ an open subset of $X$? Would anyone please tell me the answer to this question and provide suggestions or hints on how to prove it?
 A: The metric on X is induced by the norm, so that if $x = \{x_i\}$ and $y = \{y_i\}$ are elements of X, then $d(x,y) = \sup_{i \in \mathbb{N}} |x_i - y_i|$. 
Now I claim that $X_{++}$ is not an open set. 
To show a set in a metric space is open, I have to show that for any element in that set, I can find an open ball around that element entirely contained in the set. 
How could this go wrong? (This is the part that requires some thought. You have to pick the right sequence). 
Did you want me to show you a sequence in $X_{++}$ such that every open ball around it contains points of $X \setminus X_{++}$? Or is this enough of a hint so far? 
Another hint is that the sup norm is very strong. If two functions are close together in the sup norm, they are close together at every point of their domain. So if you have a sequence $\{x_i\} \in X$, an $\epsilon$-neighborhood of $\{x_i\}$ consists of sequences that differ from $\{x_i\}$ by at most $\epsilon$ at every coordinate.
Let me know if you want more hints.
