Vtyjkyuk

I wanted to prove Family of Closed set of $X$ having finite intersection property with arbitary intersection is non empty Then $X$ is compact metric space

I wanted to check my argument

On Contrary suppose $(X,d) $ is not compact space ,SO there exist open cover of X but which does not have finite subcover which can cover that X

SO $X=cup_a=1^infty G_a$ So it does not have finite subcover of this family which can cover X

That is $exists xin X $ such that $xnotin cup_a=1^n G_a $ for some n....(1)

As $G_a$ is open $to$ X $G_a=G_a'$ is closed.[Notation for complement ']

So here $cap^n_1 G'_aneq phi$ as $xincap^n_1 G'_a $ Form 1

Which implies $cap^infty_1 G'_aneq phi$ by assumption which means

$xin cap^infty_1 G'_a$ i.e $xnotin X$as covering assumption .

Now this is contradition to assumption that every $xin X$as by construction of $G_a$

How was my argument ?
Please tell me .

Your argument is muddled.

Clarify it:

We assume that $X$ has the property $ast$ that a family of closed sets with the finite intersection property has non-empty intersection.
Suppose (for a contradiction) that $X$ is not compact. Then there is
an open cover $U_i, i in I$ of $X$ that has no finite subcover.

Define $F_i = Xsetminus U_i$. Then all $F_i$ are closed.
Show that every finite intersection of $F_i$ has non-empty intersection. Hint: use that

$$cap_i in F F_i = cap_i in F (Xsetminus U_i) = X setminus (cup_i in F U_i)$$

for any (finite) subset $F$ of $I$.

By the fact that $X$ has property $ast$, we know that $cap_i F_i neq emptyset$. Show that an element of this intersection would not be covered by any $U_i$. This contradiction
then shows that the assumption that $X$ is not compact was wrong.

So $X$ is compact.