Point set topology closure

Question

Consider the following topology on $X=\{a,b,c,d,e\}$:\begin{equation}\tau=\{\emptyset,X,\{a\},\{a,b\},\{a,c,d\},\{a,b,c,d\},\{a,b,e\}\}.\end{equation} (i) List the closed subsets of $X$. (ii) Determine the closure of $\{b\}$.

Solution: A set $A$ is closed iff its complement is open. So list the complements of each set in $\tau$: \begin{equation}\emptyset,X,\{b,c,d,e\},\{c,d,e\},\{b,e\},\{e\},\{c,d\}.\end{equation} The closure $\bar{A}$ of a set $A$ is the intersection of all closed supersets of $A$. The closed supersets of $\{b\}$ are $\{b,e\},\{b,c,d,e\},X$, and their intersection is $\{b,e\}$.

My question is as follows - the author has written out the complements of the sets in $\tau$, but none of these complements lie in $\tau$ (except $\emptyset,X$), hence there are no sets in $\tau$ whose complement is open in $X$. Surely that means there are no closed elements apart from $\emptyset,X$? And for the second part, the same question applies.

Answer 1

The elements of $\tau$ are the open sets of this topology and thus the complements of the elements of $\tau$ (the sets the author listed) are the closed sets of this topology.

Answer 2

The closed sets w.r.t a topology need not to belong to the topology. Now the sets {} and X belong to T that doesn't surely mean {} and X are the only closed elements; rather that means {} and X are the sets which are both closed and open.

Answer 3

You are confusing some things here. I will try to give a detailed description of the terms used here.
First of all a topology on a set $X$ is a collection $\tau$ of subsets of $O\subseteq X$ (called open sets) such that

• $\emptyset,X\in\tau$
• Any union of open sets is itself open, that is for any family $\left(O_i\right)_{i\in I}$ with $O_i\in\tau$ for all $i\in I$ we have $\underset{i\in I}{\bigcup}O_i\in\tau$.
• Any $\textbf{finite}$ intersection of open sets is open, that is $\left(O_1,O_2\in\tau\right)\Rightarrow O_1\cup O_2\in\tau$.

We now consider the topological space $(X,\tau)$ given by $X=\left\lbrace a,b,c,d,e\right\rbrace$ and $$\tau=\{\emptyset,X,\{a\},\{a,b\},\{a,c,d\},\{a,b,c,d\},\{a,b,e\}\}.$$ $\tau$ thus contains all open subsets of $X$ and it is easy to see (but important to observe), that $\tau$ fullfills all three properties described above.
We now say that a subset $C\subseteq X$ is closed iff $\left(X\setminus C\right)\in\tau$. From this definition together with the properties of open subsets described above one obtains the following analogous properties of closed subsets.

• $\emptyset$ and $X$ are closed subsets of $X$
• Any $\textbf{finite}$ union of closed sets is itself closed, that is $\left(C_1,C_2\text{ closed }\right)\Rightarrow C_1\cup C_2\text{ closed}$
• Any intersection of closed subsets of $X$ is closed. That is for any family $\left(C_i\right)_{i\in I}$ of closed subsets of $X$ the set $\underset{i\in I}{\bigcap} C_i$ is a closed subset of $X$.\

The closed sets of our example are therefor given by $$\tau_{\text{closed}}=\{\emptyset,X,\{b,c,d,e\},\{c,d,e\},\{b,e\},\{e\},\{c,d\}\}$$ Those are now all closed subsets of $X$!
Your observation that $\emptyset$ and $X$ are the only open and closed subsets of $X$ is correct here and actually generalizes to all connected topological spaces (that is topological space that can not be written as the union of two disjoint open subsets). However this is not true in general. Consider for example the discrete topology on any space, that is you define every subset to be open, then correspondingly every subset is open and closed.
I hope this could clarify the issue.