Take the set $B=\{a\in \Bbb R^{\Bbb N}|\exists C \in \Bbb R \forall n \in \Bbb N:|a(n)|<C\}$
and the distance $d(a(n),b(n)):= \sup|a(n)-b(n)|$
Then given the subset $J:=\{a\in B| \forall n \in \Bbb N: a(n+1)\geq a(n)\}$, I want to decide if J is compact or not, given that I know it's closed ?
The theorems I think I can use :
continuous images of compact sets are compact.
any closed subset of a compact subset is closed.
Okay so from here , here is my argument:
define a function $f:[0,C] \rightarrow B$ we could just say the function is f(x)=a, where x is some value between zero and C.
clearly $[0,C]$ is compact as it is closed and bounded so by the heine borel theorem this is true.
this implies by theorem 1. that B is compact . Therefore as J is a closed subset of B , by 2. J is also compact.
Could anyone tell me if this is correct , or how I can adjust my argument ?
The set is indeed closed, but it’s closed in $\ell^\infty$ (the normed space of bounded real sequences in the supremum norm), which is itself non-compact. So it does not show compactness at all.
In fact it is very non-compact. Define for each $k \in \mathbb{N}$ the function $f_k: \mathbb{N} \to \mathbb{R}$ by $f_k(n) = 0$ for $n \le k$ and $1$ for $n>k$. These $f_k$ are all bounded non-decreasing functions in your space and one can easily check that it is a discrete and closed infinite subset of it. So your set is not even countably compact.
Another simpler argument: consider all functions/sequences that are constant. These are all in $J$ and they show that $J$ is unbounded. So it cannot be compact, as compact implies boundedness in the metric.