Vtyjkyuk

Take the set $B=exists C in Bbb R forall n in Bbb N:$

and the distance $d(a(n),b(n)):= sup|a(n)-b(n)|$

Then given the subset $J:= forall n in Bbb N: a(n+1)geq a(n)$, I want to decide if J is compact or not, given that I know it's closed ?

The theorems I think I can use :

1. continuous images of compact sets are compact.




2. 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 $k in mathbbN$ the function $f_k: mathbbN to mathbbR$ 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.