which of the following statement is true.,,,,? which of the following statement is true concerning topological spaces.?
1.continiuous  image of non compact space is non compact.
2.every metrizbale space is normal....
my attempts: option 1 is false,,,,as continuous image of compact set is compacts
             option 2 is trues  as it is given munkre topology  books

am I correct ?  pliz tell me
 A: (1) is false. E.g. Let $X$ be any non-compact space and $Y=\{0\}.$ Let $f(p)=0$ for all $p\in X.$
(2). Let $(X,d)$ be a metric space and let $A, B$ be disjoint closed subsets of $X.$ To avoid the trivial cases, suppose $A\ne \phi\ne B.$
For $a\in A$ let $d(a,B)=\inf_{b\in B}d(a,b).$ For $b\in b$ let $d(b,A)=\inf_{a\in A}d(b,A).$
For $a\in A$ we have $d(a,B)>0$ because $a\not \in \bar B.$ For $b\in B$ we have $d(b, A)>0$ because $b\not \in \bar A.$
Let $U=\cup_{a\in A}B_d(a,\frac {1}{2}d(a,B)).$ Let $V=\cup_{b\in B}B_d(b,\frac {1}{2}d(b,A)).$ 
$U$ and $V$ are open sets with $A\subset U$ and $B\subset V.$
Suppose by contradiction that $c\in U\cap V.$ Take $a\in A$ and $b\in B$ with $d(c,a)<\frac {1}{2}d(a,B)$ and $d(c,b)<\frac {1}{2}d(b,A).$ $$\text  { Then }\quad  d(a,B)\leq d(a,b)\leq d(a,c)+d(c,b)<\frac {1}{2}d(a,B)+\frac {1}{2}d(b,A)$$   $$ \text { and }\quad \;\; d(b,A)\leq d(b,a)\leq d(b,c)+d(c,a)<\frac {1}{2}d(b,A)+\frac {1}{2}d(a,B).$$ The sum of the two far-left terms in the above inequalities, which is $d(a,B)+d(b,A),$  is less than the sum of the two far-right terms, which is also $d(a,B)+d(b,A).$ This is absurd, so we conclude that $c$ does not exist.
That is, $U\cap V=\phi$.
A metric space is also a $T_1$ space. I will leave that to you.  So a metric space is a normal space.
