Thread: Definition of topology

Hello!

I got somewhat stuck when I read that in the definition of a topology on a set one may drop the first defining property.

The familiar definition reads as follows:

A Topology on a set is a collection of subsets of , called open sets, such that the empty set and the whole of are in , and is closed under finite intersections and arbitrary unions. In symbols:

(1) ;

(2) If , where $I$ is a finite index set, then ;

(3) If , where $J$ is any (finite or infinite) index set, then .

And now it is said that:

The defining property (1) may be dropped. In fact, follows, if we apply (3) to the empty family, and follows, if we apply (2) to the empty family.

Now I do see that we have to do with an instance of vacuous truth here, because whenever the index set is empty we have nothing to check.

But on this ground I could have swapped the roles, saying that:
The defining property (1) may be dropped. In fact, follows, if we apply (2) to the empty family, and follows, if we apply (3) to the empty family.
…couldn't I?

Am I wrong? Is there possibly more to this than simply addressing to vacuous truth?

Originally Posted by HAL9000

Hello!

I got somewhat stuck when I read that in the definition of a topology on a set one may drop the first defining property.

The familiar definition reads as follows:

A Topology on a set is a collection of subsets of , called open sets, such that the empty set and the whole of are in , and is closed under finite intersections and arbitrary unions. In symbols:

(1) ;

(2) If , where $I$ is a finite index set, then ;

(3) If , where $J$ is any (finite or infinite) index set, then .

And now it is said that:

The defining property (1) may be dropped. In fact, follows, if we apply (3) to the empty family, and follows, if we apply (2) to the empty family.

Now I do see that we have to do with an instance of vacuous truth here, because whenever the index set is empty we have nothing to check.

But on this ground I could have swapped the roles, saying that:
The defining property (1) may be dropped. In fact, follows, if we apply (2) to the empty family, and follows, if we apply (3) to the empty family.
…couldn't I?

Am I wrong? Is there possibly more to this than simply addressing to vacuous truth?

This is an excellent example of the very common educational method that I call "Say something

foggy, explain nothing, see the kids going nuts and go and drink a beer in full satisfaction" .

In this case the issue is far from being standard in mathematics and different people may have

different views, though I think that many (most?) people will agree with the book you're working with:

1) if we take the intersection over an (the, in fact) empty subset of , then

we can think of this as the set of all the elements that belong to all and each

of the elements (subsets of ) that appear in that intersection...but there are no

sets at all there, so we can argue that ANY element in appears there under the

logical argument "show me one subset that appears in that intersection and I'll prove you that any

element of the set is contained there"...

2) Something simmilar happens with the union of the empty family on : if

there exists an element in that union then some of the subsets there contains it...but there are no

subsets there so the union is the empty set.

Yeah, I know...not the most brilliant moment in mathematics foundations (set theory and stuff), but

if you agree with those conveniences then there's no problem.

I, for one, would never say such a thing without first trying to explain a little all this madness, and

even less would I claim that we can drop (1) because so and so, as if it were an obvious fact.

Tonio

"show me one subset that appears in that intersection and I'll prove you that any

element of the set is contained there"

I find the above statement a bit unclear. Let me try to clarify it for myself.

I think the whole matter here makes use of the ultimate premise that a priori , i.e. whenever we define a topology on X, we can be sure that every is in some open subset.

Now take the property (2). It says that the proposition " is open" is true whenever " is open" is true. But the truth of the former relies solely on the truth of , which is the same as . So in case it is true iff . But this last proposition is true a priori, because as you said, if one shows up with an open set, then we are always able to assign some into this open set, and this goes for all elements of X.

Take the property (3). Again, It says that the proposition " is open" is true whenever " is open" is true. The truth of the former depends on the truth of , which is the same as . So in case it is false. This means that we deal with all those x in X, which are in no open set. But this is a priori not fulfilled by all the elements of X. So the set of all those elements in X, for which "x in X is such that x is in no open set" is fulfilled is empty.

OK, I hope I didn't mess up the matter any further...