In the book "Introduction to Smooth Manifolds" by John M. Lee, the author defined general topology and basis as:

A topology on a set $X$ is a collection $\mathcal{T}$ of subsets of $X$, called open sets, satisfying:

  1. $X$ and $\emptyset$ are open
  2. The union of any family of open sets is open
  3. The intersection of any finite family of open sets is open

A basis for a topology on $X$ is a collection $\mathcal{B}$ of subsets of $X$ such that

  1. $X = \cup_{B \in \mathcal{B}} B$
  2. If $B_1, B_2 \in \mathcal{B}$ and $x \in B_1 \cap B_2$, there exists $B_3 \in \mathcal{B}$ such that $x \in B_3 \subset B_1 \cap B_2$

My questions are:

  • In topology definition: Why "union of any family"? Why not "union of finite elements of $\mathcal{T}$"? The same with intersection.
  • In basis definition: are those set $B$ open sets? Would it be any different if we consider the basis as a collection of general subset? And does $B_3$ have to be strictly proper subset of $B_1 \cap B_2$?
  • $\begingroup$ Note that for union, it's not the union of finitely many, it's the union of arbitrarily many elements of $\mathcal T$. For intersection, though, the demand only requires finite families. $\endgroup$ – Arthur Mar 17 '17 at 21:55

I will try to explain one point of view about what a topology is meant to represent. Hopefully it will give you a sense of why it is not too restrictive to ask for arbitrary union of open sets to be open, and why it is (sometimes) too restrictive to ask for arbitrary intersections of open sets to be open.

Let $f : \mathbb{R} \rightarrow \mathbb{R}$ be a function and consider the following statement.

The function $f$ is continuous at $x$ if $f(y)$ can be made arbitrarily close to $f(x)$ provided $y\in \mathbb{R}$ is sufficiently close to $x$.

You can think of a topology as giving a rigorous meaning to "arbitrarily close" and "sufficiently close" without refering to a distance (and even where there is no distance function).

With a topology, you can replace the statement "$P(x; y)$ holds for all $x$ sufficiently close to $y$" by "there exists an open set $\mathcal{U}\in y$ such that $P(x;y)$ holds for all $x \in \mathcal{U}$". Similarily, you can replace the statement "$x$ can be made arbitrarily close to $y$" by "$x$ can be made to belong to any open set containing $y$".

Now, consider a family $\mathcal{B}$ of open sets giving a sensible meaning the the above statements. If you think about it, you can see that the family $\tau$ obtained by taking all possible unions in $\mathcal{B}$ makes exactly the same sense out of "sufficiently close" and "arbitrarily close". Thus it is not too restrictive to allow arbitrary union of open sets to be open. However, if you allow arbitrary intersections, than you may have a problem with "arbitrarily close": an example is given by Ken in another answer.

Note: The other parts of your question are well answered by Ken.

  • $\begingroup$ Do you mean the topology on a basis $\mathcal{B}$ of topological space $(X, \mathcal{T})$ imposed by taking all union of elements of the basis $\mathcal{B}$ is exactly $\mathcal{T}$ (like basis of a vector space) ? $\endgroup$ – Vinh Khang Mar 17 '17 at 23:18
  • $\begingroup$ Yes, this is one of the things I imply. By the point (2) of the definition of a basis $\mathcal{B}$, the set of arbitrary unions of $\mathcal{B}$ is closed under finite intersection. $\endgroup$ – Olivier Mar 17 '17 at 23:21

For your second point: Yes the sets are open (at least once you generate the topology), and no $B_3$ does not have to be proper.

For the first point: Perhaps a better question would be to ask why we require the collection to be finite when talking about intersection. Let's look at the standard topology on the real line. For each $n \in \mathbb{N}$ define $$I_n = \left(-\frac{1}{n}, \frac{1}{n}\right).$$ Each of these are open sets, but $$\bigcap_{n=1}^\infty I_n= \{0\}.$$

This is not open.

  • $\begingroup$ For the second point, no, they aren't open; not yet. They are any subsets as long as they fulfill the given requirement. Then, you may use that basis to generate a topology (the topology will consist of all possible unions of elements of $\mathcal B$), and once you've done that, the $B_i$ will be part of that topology, and as such, open. $\endgroup$ – Arthur Mar 17 '17 at 22:03
  • $\begingroup$ @Arthur Fair point, I was thinking of the end result. $\endgroup$ – Ken Duna Mar 17 '17 at 22:04

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