I know "every normal space is regular space". But today, while practicing some problem I saw the following,
Let $X=\{a, b,c\}$ and $τ=\{ ∅, \{a\}, \{b\}, \{a,b\}, X\}$ be a topology on $X$.
Clearly I can see, for $a∈X$ and closed set $\{c\}$ there does not exists disjoint open sets $U$, $V$ such that $a∈U$ and $\{c\}⊆V$ and hence $X$ is not regular. So it must be not normal!
But what I saw in key that, $X$ is normal! How it's possible? How $X$ is normal? Please help me...
On
If NORMAL and REGULAR are both defined without requiring T1, then normal does NOT imply regular (and, of course, vice versa). The above example works, as does any nontrivial space whose open sets are a chain under containment: There are no disjoint closed sets OR open sets. The former makes the space vacuously normal; but regularity fails due the existence of a closed set and an excluded point.
In your space $X$ the closed sets are $\{X, \{b,c\}, \{a,c\}, \{c\},\emptyset\}$ (the complements of the open sets).
This means there are no pairs of non-empty closed disjoint sets. So voidly (there is nothing to do) $X$ is normal.
But as you rightly say you cannot separate the point $a$ from the closed set $\{c\}$ (as the only open set containing $c$ is $X$, e.g.), so $X$ is not regular.
It's common to have a combination of $T_1$ with either normality or regularity. This is done, because then, as $T_1$ is equivalent to the fact that all singletons $\{x\}$ are closed, we are sure that there are actually lots of pairs of non-empty disjoint closed sets and pairs of points and closed sets such that we can apply the regularity and normality. In particular, in both cases we know that $X$ is Hausdorff: separating points $x \neq y$ can also be seen as separating the closed sets $\{x\}$ and $\{y\}$ or the point $x$ and the closed set $\{y\}$, say.
If we have $X$ that $T_1$ and normal, then we can separate all points $x\in X$ and $C \subseteq X$ closed with $x \notin C$, because we just separate the pair of (disjoint) closed sets $\{x\}$ and $C$ by normality and use those open sets. So then $X$ is $T_1$ and regular.
If you want to be really subtle, note that $X$ $T_0$ and regular already implies that $X$ is $T_1$ (and thus $T_2$ as we saw): let $x \neq y$. By $T_0$ we have that there either is an open subset $O$ such that $x \in O, y \notin O$ or the other way round. First case: define $C= X\setminus O$ which is closed such that $x \notin C$, so regularity gives us that there are disjoint open sets $U_x$ and $V$ such that $x \in U_x$ and $y \in X\setminus O \subseteq V$. But then $V$ is an open subset that contains $y$ but not $x$, so completing the proof of $T_1$-ness. The second case is similar, interchanging the roles of $x$ and $y$. Note that normal plus $T_0$ need not imply $T_1$ as witnessed by the Sierpinski space, e.g.
The combination of $X$ being regular and $T_1$ (or $T_0$) is usually called $T_3$ (the $T$ stands for German "Trennungsaxiom" (separation axiom) and for normality plus $T_1$ we have $T_4$ as a shorthand. Beware that some texts use $T_3$ for just plain regular, etc. Always check such things to avoid confusion. Or the combination is called "regular" etc. I'm in favour of using the $T_3, T_4$ for the strengthened versions, because then $T_j \Rightarrow T_i$ when $j > i$.