Standard topology is coarser than lower limit topology?

[PeroK]

You do not allow it, you demand it! And that makes it equivalent to $X\in \tau$ only under the shell.

I am absolutely sure that we wouldn't have $\emptyset\, , \,X \in \tau$ as an axiom if it wasn't necessary. Pawel Samuilowitsch Urysohn, Ascher Zaritsky, Felix Hausdorff, Andrej Nikolaevic Tichonov, and Hans Julius Zassenhaus would have long dropped it!

Well, it's clearer with a separate axiom than relying on an ugly and somewhat artificial interpretation of unions and intersections. And, as pointed out in an earlier post, it doesn't save any practical effort in checking that a topology satisfies the axioms.

 

[WWGD]

You may find it a ridiculous topology, but $\{\emptyset, \{a\}, X\}$ is a perfectly good topology while $\{\{a\}\}$ is not. There is no requirement that any open set except the full set contains any particular element. Since the full set is included by axiom, that each point has a neighbourhood becomes trivially true and uninteresting. There is no requirement for a topology that the full set should result from the union of any set of open sets that does not include the full set. In fact, any set $X$ can be equipped with the trivial topology $\{\emptyset,X\}$ where it is obvious that you cannot obtain $X$ from any union of open sets that does not already include $X$. This is a matter of fact, not opinon, although you are free to dislike the fact.

Edit: I never said a particular element should be contained in a particular set. Both the discrete and indiscrete topologies are used for little but counterexamples . In the Discrete topllogy on X, every map out of X is continuous. In the Indiscrete, elements in X can't be separated or distinguished. Not much useful comes out of either.

 

[fresh_42]

You may find it a ridiculous topology, but $\{\emptyset, \{a\}, X\}$ is a perfectly good topology while $\{\{a\}\}$ is not.

I know, that was my example. And $\{\{a\}\}$ isn't a topology on $\{a,b\}$ even if we included $\{\emptyset,\{a\}\}.$

I find the request ridiculous that every element has to be part of some open set, neither the topology nor the persons who make that claim. I am saying that this is of equal quality as $\{a,b\}\in \tau$ is, and only disguises this fact. It complicates things by hiding the entire axiom behind notationally induced logic. One has to add the element $b$ beforehand ónly to shout "Eureka!" afterward when it comes out of the cylinder. I find this magic trick ridiculous, because that's what it is.

 

[fresh_42]

I never said a particular element should be contained in a particular set.

You did implicitly. Without such a statement you will have no chance to prove that $\tau:=\{\emptyset\, , \,\{a\}\}$ is no topology on $X:=\{a,b\}.$ As I said before, there is no way to create $X$ as a union of sets from $\tau.$ You have to make an assumption that is basically equivalent to $X\in \tau,$ only hidden in the right shell of the shell gamer.

 

[WWGD]

You did implicitly. Without such a statement you will have no chance to prove that $\tau:=\{\emptyset\, , \,\{a\}\}$ is no topology on $X:=\{a,b\}.$ As I said before, there is no way to create $X$ as a union of sets from $\tau.$ You have to make an assumption that is basically equivalent to $X\in \tau,$ only hidden in the right shell of the shell gamer.

I wish you didn't have to hurl accusations like shell gamer. I guess I should have stood by my decision to stay out, while standing by what I've said.

 

[fresh_42]

I wish you didn't have to hurl accusations like shell gamer. I guess I should have stood by my decision to stay out, while standing by what I've said.

That is my impression. The axiom is hidden beneath the notation such that it gets invisible, only to say it is there at the end of the argument. I had this association from the stackexchange post @PeroK has linked to. They hid it under $X =\cup \tau.$ This is only another way to say that every element has to be in an open set, which is another way to say $X\in \tau.$

The shell game association came from

which looked to me like hiding $X\in \tau$ in the notation $\tau$ is a topological space, which seems to me a ταυτολογία.But to make it clear: By no means I did ever suggest that you are a shell gamer or even have any similarities with one. The minimal system of axioms actually works for the topologies as defined in post #1. They do not work for the example I gave. So either we drop the requirement that a definition has to cover all cases, or we will have to use tricks like the one on stackexchange.

I mean, read it again dear @WWGD! stackexchange and I sould not associate shell games?

 

[Orodruin]

Edit: I never said a particular element should be contained in a particular set.

You kind of did, at least you make it seem like it.

This

How about defining your topology on the Reals themselves with just the empty set and {a}?

is not a topology on $\mathbb R$ because $\mathbb R$ is not in the topology so what follows simply does not apply. However, if you do include $\mathbb R$ then it is a topology and the statement that every $x \in \mathbb R$ must be in an open set is trivial because $x \in \mathbb R$, which is open.

 

[WWGD]

Thanks for your response. But this still not clear for me, I wish there is a rigorous approach for your statement. Suppose that $\delta=1$, we have
$$[x,x+1)\subset(x-1,x+1).$$
I'm not sure that this is still true.

A standard argument/technique is to show that every element on the left hand is contained in the set on the right hand.

 

[WWGD]

You kind of did, at least you make it seem like it.

This

is not a topology on $\mathbb R$ because $\mathbb R$ is not in the topology so what follows simply does not apply. However, if you do include $\mathbb R$ then it is a topology and the statement that every $x \in \mathbb R$ must be in an open set is trivial because $x \in \mathbb R$, which is open.

That was the whole point. A topology defined on a set X must be able to talk about anything in X. Here {a} stood for the set in which the topology is defined, where {a} is any Real number.