# Thread: Topological space: Condition 3

1. ## Topological space: Condition 3

$\for A,B \in \mathfrak{T}, A \cap B \in\mathfrak{T}$

What is a good way to prove this true? One video I saw said it can be proved by induction but I haven't done an induction proof on a set before. I thought maybe an element proof would be easier to work out.

2. ## Re: Topological space: Condition 3

Choose an element \displaystyle \begin{align*} x \end{align*}. If \displaystyle \begin{align*} x \in A \end{align*} and \displaystyle \begin{align*} A \subset \mathcal{I} \end{align*} then \displaystyle \begin{align*} x \in \mathcal{I} \end{align*}.

But if \displaystyle \begin{align*} x \in B \end{align*} and \displaystyle \begin{align*} B \subset \mathcal{I} \end{align*}, then \displaystyle \begin{align*} x \in \mathcal{I} \end{align*}.

Since x is in both A and B, that means \displaystyle \begin{align*} x \in A \cap B \end{align*}, and since we know \displaystyle \begin{align*} x \in \mathcal{I} \end{align*}, that means \displaystyle \begin{align*} A \cap B \subset \mathcal{I} \end{align*}.

3. ## Re: Topological space: Condition 3

How about using the basic definitions? " $A\subseteq X$ is defined as "if $x\in A$ then $x\in X$.

Suppose $x\in A\cap B$ then, by the definition of $A\cap B$, $x\in A$. Since $A\subset T$, ....

4. ## Re: Topological space: Condition 3

Both of those are suitable, theres no leap to see its true.

Thinking about the idea of proving it by some sort of induction is interesting though

would that be along the lines

for two such sets

show its true for two sets intersecting

then assume K sets intersecting $A \cap B \cap......\cap K$ is true

then proving K+1 sets having an intersection are in $\mathfrak{T}$

$A \cap B \cap......\cap K \cap K+1$

This seems maybe incorrect cause what is the K+1 set, the Kth set plus an added element or a set not equal to the Kth set but having intersection with the Kth set that is nonempty? (does it even matter if the intersection is non-empty considering the empty set is in $\mathfrak{T}$ ? )

5. ## Re: Topological space: Condition 3

If I may ask, what is $\mathfrak{T}$?

If it is a topology, then it cannot be proven, it is axiomatically true. If it is some collection of sets, whether or not it is true depends on the collection.

Both ProveIt's response and HallsOfIvy's response seem to indicate that they believe $A,B$ are subsets of $\mathfrak{T}$. I am not sure why they believe this to be true.

6. ## Re: Topological space: Condition 3

Originally Posted by Deveno
If I may ask, what is $\mathfrak{T}$?

If it is a topology, then it cannot be proven, it is axiomatically true. If it is some collection of sets, whether or not it is true depends on the collection.

Both ProveIt's response and HallsOfIvy's response seem to indicate that they believe $A,B$ are subsets of $\mathfrak{T}$. I am not sure why they believe this to be true.
yes it is a topology, a book I am reading on point-set topology recommended that it would be a good idea to prove it. I also watched a lecture on point-set topology that recommended proving it to oneself--stating that using an induction proof would be useful.

So why do you say it cannot be proven?

7. ## Re: Topological space: Condition 3

If $\mathfrak{T}$ is some SPECIFIC topology, then it could be proven, but we would have to know what defines $\mathfrak{T}$.

A topology on a set $X$ is, by definition:

1. A subset of $\mathcal{P}(X)$ closed under finite intersections and arbitrary unions.
2. Such a subset that contains $\emptyset$ and $X$ as elements.

The elements of $\mathfrak{T}$ are called "open sets".

Alternate definitions do exist, some texts define "neighborhood systems" or "closed sets" instead.

8. ## Re: Topological space: Condition 3

Originally Posted by Deveno
If $\mathfrak{T}$ is some SPECIFIC topology, then it could be proven, but we would have to know what defines $\mathfrak{T}$.

A topology on a set $X$ is, by definition:

1. A subset of $\mathcal{P}(X)$ closed under finite intersections and arbitrary unions.
2. Such a subset that contains $\emptyset$ and $X$ as elements.

The elements of $\mathfrak{T}$ are called "open sets".

Alternate definitions do exist, some texts define "neighborhood systems" or "closed sets" instead.
My book also contains a 3rd condition

Which is written as $U \cap V \in \mathfrak{T}$ whenever $U,V \in \mathfrak{T}$

and the lecture listed it as $\for A,B \in \mathfrak{T}, A \cap B \in\mathfrak{T}$

but the intersection and union are covered in your first condition as opposed to my book which has them separately as the second and third conditions. So that makes sense

9. ## Re: Topological space: Condition 3

Originally Posted by Jonroberts74
My book also contains a 3rd condition

Which is written as $U \cap V \in \mathfrak{T}$ whenever $U,V \in \mathfrak{T}$

and the lecture listed it as $\for A,B \in \mathfrak{T}, A \cap B \in\mathfrak{T}$

but the intersection and union are covered in your first condition as opposed to my book which has them separately as the second and third conditions. So that makes sense
OK, but my point is this: you don't "prove these are true", you prove instead that they are true for some defined collection $\mathfrak{T}$.

For example, the set of all unions of open intervals form a topology for the real line.

If we take an arbitrary union of unions of open intervals, it's just a larger union, so is still in the collection.

On the other hand, the intersection of any two open intervals is again a (possibly empty) open interval, and the intersection of two sets which are each unions of open intervals, is a union of the intersections.

We can write the entire real line as: $\displaystyle \bigcup_{n \in \Bbb Z} (n-1,n+1)$, so it is in the set, and we can write $\emptyset = (a,a)$ for any real number $a$, which shows these two sets are in our collection.