True or False Questions about Open , Closed, and Closure Set. 
*

*For any set A, complement of closure A is open => True since closure A is closed and complement must be open.

*If a set A has an isolated point, it cannot be an open set.  => I think it is true, but can not think of proof.
3.Set A is closed if and only if A= closure of A => True by following definition.


*If A is a bounded set, then s=supA is a limit point of A => False Let A={1,2,3} then supA is 3 but 3 is not a limit point. 

*Every finite set is closed => true, singleton element is closed and finite union of closed sets is closed.

*An open set that contains every rational number must necessarily be all of R
I think it is true, but anyone can explain? 
If there are any wrong answers, please point out and show me some counterexmaples.
Thanks 
 A: *

*Looks good.

*This is true in $\mathbb{R}$, but not in general metric spaces. For a trivial example, consider the discrete metric on a single point; for a slightly less trivial example, consider the set $\{1\}\subset\{1/n:n\in\mathbb{Z}^+\}$ with the metric induced from $\mathbb{R}$.

*Looks good.

*Looks good.

*True in metric spaces but not in topological spaces, for instance by taking the indiscrete topology.

*False, consider $(-\infty,\sqrt{2})\cup(\sqrt{2},\infty)$. There are stranger examples; for instance, if we enumerate $\mathbb{Q}=(q_i)_{i\in\mathbb{N}}$, then $\bigcup (q_i-2^{-i},q_i+2^{-1})$ is open and contains the rationals, but the sum of the lengths of these intervals is merely $4$!
A: The answer to the six question is false because $\mathbb{Q}$ is a totally disconnected subspace of $\mathbb{R}$. In particoular there isn't the Dedekind's axiom for the rational numbers. For example take the open set $\lbrace r\in \mathbb{R}:r<\sqrt{2} \rbrace\cup\lbrace r\in\mathbb{R}:r>\sqrt{2} \rbrace\supseteq \mathbb{Q} $. This is open because is union of open sets of the real line.
