# Why is the set of all non-decreasing sequence not open in the real sequence space?

Given the set $B=\{a\in \Bbb R^{\Bbb N}|\exists C \in \Bbb R \forall n \in \Bbb N:|a(n)<C\}$ with the distance $d(a(n),b(n)):= \sup|a(n)-b(n)|$.

Take the set $J:=\{a\in B| \forall n \in \Bbb N: a(n+1)\geq a(n)\}$

I know that $J$ is not open in the metric space $B$ but I want to make sure I know exactly why.

So take the null sequence $0\in J$. This is an element of J as each member is zero and so the condition $a(n+1)\geq a(n)$ is satisfied as zero is equal to zero.

Next consider an open ball centered at $0$, $B_r(0)$, so the taking the metric $d(0,b(n)):=\sup|0-b(n)|= \sup |b(n)|$ the open ball is $B_r(0)=\{b(n)|d(0,b(n)) <r\}$, In other words $\sup|b(n)|<r$. okay honestly after this I'm stuck. I don't know why no open ball exists in $J$ which is centered at zero . could anyone explain ?

The assertion is equivalent to the existence of arbitrarily small points in the complement of the set. Now consider the element $$c(n) := \begin{cases} \epsilon & n=1 \\ 0 & n \ge 2. \end{cases}$$ It violates the condition of $J$, but is close to zero with regard t your metric.
• okay I think I understand...c(n) is arbitrarily close to my null sequence 0 so it exists in an open ball $B_\epsilon(0)$ but it does not exist in Jso then this open ball does not exist soley in J and so J cannot be open ? – Voltron Aug 2 '18 at 14:06