How to proof that any sequence in $A\in R^n$ has limit also in A if A is closed?
I am interesting in proof without referering to the definition of closed set as a set where any sequences limit is also in closed set. Conversly, I am interested in opposit logic: how the fact that the set is closed guarantee that any sequence will have limit also lying in this set.
Will be very greatful for help!
It depends on what you take as the definition of closed. If we take the definition below, the claim is trivial.
Let $(a_{n})_{n}$ be a sequence in $A$ converging to $a$. That is, for each neighborhood $V$ of $a$, there exists $N$ such that for all $n\geq N$, $a_{n}$ is in $V$. Can you show that $a$ is a limit point of $A$?
If instead we take the definition below, a bit more work is required.
Let $A$ be a closed set and $(a_{n})_{n}$ a sequence in $A$ converging to some $a$. By the definition above, there exists some open set $B$ such that $A^{c}=B$. To arrive at a contradiction, suppose $a$ is in $B$. Since $B$ is a neighbourhood of $a$, it follows that the sequence $(a_n)_n$ is eventually in $B$. However, this is a contradiction, since it was assumed that the sequence $(a_{n})_{n}$ takes values in $A=B^c$.