let $(X,d)$ be a metric space
1) I have to prove $X$ is closed. this can be very easily proved by using the fact $X-X=\emptyset$ is open. but an alternate approach can be made by proving $X$ is closed if $\partial X \subseteq X$ where $\partial X$ is the set of all boundary points of $X$. so when I try to prove this it seems like $\partial X=\emptyset$. so my problem is
Is boundary of $X=\emptyset$
my argument is this,
first, let us consider the definition of a boundary point
$x_0$ is a boundary point of $S$ if $\forall r>0: B(x_0,r)\cap S\neq \emptyset$ and $\forall r>0: B(x_0,r)\cap (X-S)\neq \emptyset$
now let $x\in X$ and $r \in \mathbb R$.
thus $B(x,r)\cap X\neq \emptyset$ since $x\in X$
but $B(x_0,r)\cap (X-X)= \emptyset$ since $(X-X)= \emptyset$
thus $\forall x \in X$, $x$ is not a boundary point. therefore boundary of $X=\emptyset$
.............................................................................
2)and I have another problem, in Simmons introduction to topology and modern analysis book when it tries to prove any finite intersection of open sets is open it states that
"$G=\cap_iG_i$ where $\{ G_i \}$ is an arbitrary class of open sets. if ${G_i}$ is empty then$G=X$ i cannot figure it out how, when ${G_i}$ is empty $G=X$. can someone help me out here?
The boundary of a set $A$ is $\overline{A} \cap \overline{X \setminus A}$ : every neighbourhood (or ball if you prefer) of a boundary point intersects $A$ and also its complement. So the boundary of $X$ is $\emptyset$, as there are no points in the complement to intersect. This is also the essence of your argument.
The last point is subtle, and involves the intersection of an empty family of sets, which is discussed in this short video and shown that this equals the whole set, so $X$ in your case. This has to do with the fact that $\forall x \in A: \phi(x)$ is a true statement in logic when $A = \emptyset$, see Wikipedia.