Determine whether a property possessed by every term in a convergent sequence is necessarily inherited by the limit. I'm having difficulty coming up with actual sequences that have the properties below. I've included my thoughts on the questions below.
Assume that $(a_n)\rightarrow a$.


*

*If every $a_n$ is an upper bound for a set $B$, then $a$ is also an upper bound for $B$


Is it possible to construct a strictly decreasing sequence, in which all of the terms are upper bounds, but the limit point is less than the supremum of $B$?


*If every $a_n$ is the complement of the interval $(0,1)$, then $a$ is also in 
the complement of $(0,1).$


I'm actually not sure what to consider at all here. 
 A: *

*Weak inequalities are preserved in the limit, so if for any $b \in B$, $a_n \geq b$ for all $n$, then $a \geq b$. 


An way to see that $a\geq \sup B$ is to suppose that $\sup B > a$. Then, we could find a neighborhood around $a$ such that the entire neighborhood is below $\sup B$. Since $a$ is a limit point, there must be an $a_n$ in this neighborhood, but then $a_n$ would not be an upper bound of $B$, since it is less than the least upper bound. 


*$(0,1)$ is an open set, so $(0,1)^c$ is closed. This means that any convergent sequence in $(0,1)^c$ converges to a point in $(0,1)^c$. 


Suppose that $a \in (0,1)$. Then, pick some $0<\epsilon<\max(|a-0|, |a-1|)$. The ball of radius $\epsilon$ must lie entirely in $(0,1)$, and there must be a point in the sequence of $a_n$'s that lies in this ball, because $a$ is a limit point. So there is a point in the sequence that does not lie in $(0,1)^c$, which is a contradiction. 
A: *

*Let the supB = b  and since every term of the sequence is an
upper bound of B so, it implies that b –an ≤0.
Then lim(b–an) ≤ 0
i.e, b–a ≤ 0
i.e. b ≤ a
So, a is an upper bound for B.

