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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 ?

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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.

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  • $\begingroup$ 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 ? $\endgroup$ – excalibirr Aug 2 '18 at 14:06
  • $\begingroup$ could you tell me if that's correct ? $\endgroup$ – excalibirr Aug 2 '18 at 14:13
  • $\begingroup$ Unfortunately my browser does not render comments with math-dollars. Could you rephrase accordingly? $\endgroup$ – AlgebraicsAnonymous Aug 2 '18 at 14:15
  • $\begingroup$ okay I think I understand...c(n) is arbitrarily close to my null sequence 0 so it exists in an open ball Bϵ(0) but it does not exist in Jso then this open ball does not exist soley in J and so J cannot be open ?I took out the math dollars around the open ball... i thjink this is what you meant $\endgroup$ – excalibirr Aug 2 '18 at 14:16
  • $\begingroup$ if not I can rewrite it again, just not sure if that's what you meant or not $\endgroup$ – excalibirr Aug 2 '18 at 14:21

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