Unnecessary property in definition of topological space A set $X$ with a subset $\tau\subset \mathcal{P}(X)$ is called a topological space if:


*

*$X\in\tau$ and $\emptyset\in \tau$.

*Let $L$ be any set. If $\{A_\lambda\}_{\lambda\in L}=\mathcal{A}\subset\tau$ then $\bigcup_{\lambda\in L} A_\lambda\in\tau$.

*Let $M$ be finite set. If $\{A_\lambda\}_{\lambda\in M}=\mathcal{A}\subset\tau$ then $\bigcap_{\lambda\in M} A_\lambda\in\tau$.


Let $\emptyset=\mathcal{A}=\{A_\lambda\}_{\lambda\in N}$, i.e $N=\emptyset$. Then by 2:
$$\bigcap_{\lambda\in N} A_\lambda=\{x\in X; \forall \lambda\in N\text{ we have }x\in A_\lambda\}=X\in\tau,$$
since $N$ is empty. And by 3:
$$\bigcup_{\lambda\in N} A_\lambda=\{x\in X; \exists \lambda\in N\text{ such that }x\in A_\lambda\}=\emptyset\in\tau,$$
since $N$ is empty. Then 2 and 3 implies 1.
Many books define a topology with 1,2 and 3. But I think that 1 is not necessary because I was prove that 2,3 $\Rightarrow$ 1.
Am I right?
 A: Yes, you’re right. However, it’s more convenient to include (1). First, the fact that $\varnothing$ and $X$ are both open sets is important enough to be worth emphasizing. Secondly, many people are a bit uncomfortable with the union or intersection of an empty collection, just as many are a bit uncomfortable with the sum or product of the empty collection of real numbers. If you include (1) as part of the definition, you don’t have to deal with this awkwardness. This is especially important when you’re teaching elementary topology: most students at that stage definitely have trouble with the union and intersection of an empty collection.
Correction: I accidentally inverted one negation mentally when I thought about this the first time. You’re right about $X$, but not about $\varnothing$. Let $S=\{x\in X:\forall A\in\varnothing(x\in A)\}$; then $S$ is actually $X$, not $\varnothing$. To see this informally, ask yourself how there could be an $x\in X\setminus S$: there would have to be an $x\in X$ such that $\exists A\in\varnothing(x\notin A)$. But there isn’t any $A\in\varnothing$, so there is no such $x$, and $S=X$. (And I see now that Asaf has already pointed this out in his answer.)
A: I don't know if there's still interest in this thread, but I'm going to answer anyway.
As Asaf points out, $\bigcap \emptyset$ may not exist, and it certainly doesn't equal $X$. The problem occurs because subsets don't remember the larger set they were cut from. However, this is a quirk of how subsets are defined. And, the problem can be avoided by redefining the meaning of "subset." This will allow us to rewrite the definition of a topological space.
Firstly, lets agree that by "function", we will mean an ordered triple $(f,X,Y)$. Then we can define that a subset of $X$ is a function $A : X \rightarrow 2,$ where $2 = \{0,1\}.$
Now for a bit of notation. As shorthand for $A(x)=1$, lets write $x \propto A,$ which can be read "$x$ is an element of $A$."
Lets also write $A \diamond X$ to mean that $A$ is a subset of $X$. Note that the subset relation is no longer transitive, so we also need a containment relation. So if $A,B \diamond X$, lets write $A \subseteq B$ in order to mean that for all $x \in X$ it holds that if $x \propto A$, then $x \propto B$.
Also, lets write $2^X$ for the powerset of $X$. Formally, we define $$2^X := \{A : X \rightarrow 2 | A \mbox{ is a function}\}.$$
Thus $A \diamond X$ if and only if $A \in 2^X$.
Finally, lets write $A = \{x \in X | P(x)\}$ in order to mean that $A \diamond X$, and that $x \propto A$ iff $P(x)$.
Given these conventions, we can define the intersection of $\mathcal{A} \diamond 2^X$ as follows.
$$\bigcap \mathcal{A} = \{x \in X|\forall A \propto \mathcal{A} : x \propto A\}$$
Unions can be defined similarly.
Finally, the payoff: letting $\bot$ denote the least subset of $2^X$, and letting $\top$ denote the greatest subset of $X$, we see that
$$\bigcap \bot = \top.$$
We're now in a position to rewrite the definition of a topological space.

A set $X$ with a subset $\tau \diamond 2^X$ is called a topological space if:


*

*For all $\mathcal{A} \subseteq \tau$ it holds that $\bigcup \mathcal{A} \propto \tau$.

*For all finite $\mathcal{B} \subseteq \tau$ it holds that $\bigcap \mathcal{B} \propto \tau$.
A: The problem is that as you formulated that $\bigcap\varnothing$ is the set of all elements $x\in X$ that for every $A\in\varnothing$ we have $x\in A$, this is satisfied by all the elements of $X$.
Note that $\bigcup\varnothing$ is well-defined in ZF since the axiom of union says it is a set, and we can prove that this set is indeed $\varnothing$. However $\bigcap\varnothing$ is not well defined, because as I remark above, it can result with the collection of "everything", which in set theory is not a set at all, and in this case - not even empty.
