Given sequence $(y_n)_{n \in N}$ of real numbers with exactly 2 accumulation points being $\sqrt{2}$ and $\sqrt{3}$.
a)Check if the set A={$\sqrt{2},\sqrt{3},y_1,y_2,...$} is compact
b)Check if the set B={$\sqrt{2},y_1,y_2,...$} is totally bounded
Any help is valued since i am not sure of how to solve it.
My solution so far(Probably Wrong)
Check if the set A={$\sqrt{2},\sqrt{3},y_1,y_2,...$} is compact
As long as we show that A is closed and bounded then it should be compact.
We know that both extreme points are included in the set then the set A is bounded by those proving one of the conditions right.
Since $(y_n)$ is a sequence that belongs in real numbers then it is close and open. In this case we will use the close set. As for the extreme points as a separate set we know since the number of elements in that set is finite then we know that the set is compact by itself thus closed.
Given those 2 factors we can prove that the set is compact
b)Check if the set B={$\sqrt{2},y_1,y_2,...$} is totally bounded
A totally bounded set can be covered by finitely many subsets of every fixed "size" (where the meaning of "size" depends on the structure of the ambient space.) Since we have n elements in $(y_n)$ (meaning finite) then we can separate the set into smaller equal subsets depending on the structure of the ambient space The addition extreme point $\sqrt{2}$ makes this equal distribution of those subsets unequal meaning its not totally bounded.
PS. English is not my first language i hope everything in my solution is understood.