Confusion regarding the definition of topology space, closed set.

Question

Definition of topology space

A topology $\mathcal T$ on a set $X$ is a collection of subsets of $X$ such that i) $\emptyset,X\in\mathcal T$ ii)if $O_\alpha\in\mathcal T$ for each $\alpha\in A$, then $\bigcup_{\alpha\in A}O_\alpha \in \mathcal T$. iii) if $O_1,...,O_n \in \mathcal T$, then $O_1\cap...\cap O_n\in \mathcal T$. A set $O\subseteq X$ is called open if $O\in \mathcal T$. The pair $(X,\mathcal T)$ is called topological space.

Basically any set in the topological space is open by definition. The textbook I am reading also lists out the way to define topological space by closed sets. I understand that part. In the textbook, then closed set is defined as

A subsef $F$ in $(X,\mathcal T)$ is called closed if $X-F$is open, that is, $X-F\in \mathcal T$

So if a set is closed, then it must be clopen in the topological space, am i right? There is not set which is neither closed or open in any topological space? It doesn't sound right. Does the textbook have a typo?

Answer 1

The textbook is correct, you've just misinterpreted what it's saying. The textbook says the topology is a collection of subsets of $X$. This does not have to be the collection of all subsets of $X$. It just has to be a collection of subsets of $X$ that satisfies the axioms.

For example, consider the trivial topology on $X$. In this topology, the only open sets are $X$ and $\emptyset$. This obviously satisfies axiom 1. $\emptyset\cup X=X$, so it satisfies axiom 2. Finally, $\emptyset\cap X=\emptyset$, so it satisfies axiom 3. So, this is indeed a topology. Crucially, any non-empty $A\subsetneq X$ is not open in this topology. Since $A\neq\emptyset$, $A^c\neq X$ so $A$ is not closed either.

Answer 2

"Basically any set in the topological space is open by definition."

No. A set is open iff it happens to be a member of the topology $\mathcal{T}$ which is part of the definition of the space. The definition just says that at least two sets are in $\mathcal{T}$ and if we have some sets in it, then their union and all finite intersectionn must also be in it. So $\mathcal{T}$ obeys some properties and is not at all arbitrary.

A set $F$ is closed if $X-F$ (or $X\setminus F$ or $F^\complement$ in different notations) is open so is in the collection $\mathcal{T}$.

We could define a specific topology on $X$ by $X=\mathscr{P}(X)$, the power set, and it will obey the axioms. In that case only can we say that all subsets are open and then also all subsets are closed by the definition of closed. This is quite a special case.

Answer 3

First, we cannot say any set in a topological space is open unless the topology is the discrete topology.

for instance, consider $\mathbb R$ with usual topology. then any singleton set $\{a\}$ is not open. So in general there sets in topological space which may not be open.

Secondly, if a set is closed then it is open is also false in general. A set $F$ is closed if its complement is open. to be clopen the set must also be open.

finally, there are sets which are neither open nor closed. for example $\mathbb R$ with usual topology then any right open left closed interval is not open also not closed.