Standard topology is coarser than lower limit topology?

[WWGD]

I think this subject is more than discussed in detail.

post #2 exhibits the error in the OP's version
post #6 contains the entire proof (up to the quantification of $x$, the arbitrariness of $U$, etc.)
post #14 clarifies the confusion about basis sets
post #19 clarifies that $\emptyset\, , \,\mathbb{R}$ are also covered

All other posts could easily be ignored, deleted, or considered noise.

Just saying, in case anyone wants to read only what is essential.

This is a matter of just reading statements in post #1. The topologies are _ explicitly _ given as such, and nowhere is it asked to verify that these are actual topologies. I never offered a proof that these were topologies, but just provided additional verification.

 

[Orodruin]

This is a matter of just reading statements in post #1. The topologies are _ explicitly _ given as such, and nowhere is it asked to verify that these are actual topologies. I never offered a proof that these were topologies, but just provided additional verification.

The point is that the topologies are complete as given and what is given is not just a base for the topologies. There is no need to go around making intersections.

Yes, the topologies are given but it is still a relevant question if the given open sets are the full topology just by the given definition or not. If you want to do that you first check that the full set and the empty set are included, then you check that all (finite) intersections and (finite and infinite) unions are in the topology. In the verification that $\tau_1$ is a topology, $[0,1)\cap[1,3) = \emptyset \in \tau_1$ is telling us that this intersection is compatible with the intersection of open sets being open, not that $\emptyset \in \tau_1$, which is clear already from the definition of $\tau_1$ as it should be. If $\emptyset$ was not included in the definition, $\tau_1$ would fail both the check that it should be and the check that finite intersections of open sets are open.

When checking if $T$ is a topology or not, saying that a particular set is in $T$ because it is an intersection of sets in $T$ presupposes the property of a topology that any finite intersection of open sets is open and therefore not very useful for determining if the set belongs to $T$ or not. You can of course sometimes reach the conclusion that $T$ is not a topology unless a particular set is also included. Removing the empty set from the topologies above would break both the requirements of the empty set being included and of finite intersections of open sets being open.

Take another topology, where removing the empty set would have still have passed the checks of unions and intersections, as an additional example:
$$
\tau_2 =\{\emptyset, \mathbb R, (a, \infty): a\in \mathbb R\}.
$$
The verification that this is a topology on $\mathbb R$ goes:

1. $\emptyset \in \tau_2$ and $\mathbb R \in \tau_2$. Ok.

2. Any union
$$
U = \bigcup_i (a_i,\infty) = (\inf_i a_i, \infty).
$$
It is clear that this is in the topology when the infinum is finite and when there is no lower bound $U = \mathbb R$, which is also in the topology. The union of $\mathbb R$ with any other set is $\mathbb R$ so this is also in the topology. The union of $\emptyset$ with $U$ is $U$, which is in the topology if $U$ is. Hence, any union of open sets is open. Ok.

3. The finite intersection
$$
\bigcap_i (b_i, \infty) = (\max_i b_i,\infty) \in \tau_2.
$$
Furthermore, for any set $U$ we have $\mathbb R \cap U = U$ which is in the topology if $U$ is and $\emptyset\cap U = \emptyset$, which is in the topology. Hence, any finite intersection is in the topology. Ok.

Specifically note that the empty set needed to be included explicitly in the definition of the topology. If it were not, 2 and 3 would still hold but 1 would not. There is no intersection of the other sets that is the empty set.

Ok, it is late and I am suffering from insomnia. This probably became more of an incoherent ramble than I intended.

 

[PeroK]

Yes, the topologies are given but it is still a relevant question if the given open sets are the full topology just by the given definition or not. If you want to do that you first check that the full set and the empty set are included, then you check that all (finite) intersections and (finite and infinite) unions are in the topology.

My point, for what it's worth, is that you can drop the first axiom about the empty set and full set being in the topology - as long as you recognise the empty intersection and empty union in the other axioms.

It doesn't practically change what you have to test, since checking the empty intersection and empty union are special cases that amount to checking that the empty set and full set are in the topology! But, you can save yourself an axiom if you want.

 

[fresh_42]

I am almost certain that you meant this ironically. But for all readers who drop by ...

It doesn't practically change what you have to test, since checking the empty intersection and empty union are special cases that amount to checking that the empty set and full set are in the topology! But, you can save yourself an axiom if you want.

Would be an ugly way to see it as it only applies to certain cases like the given one here. One cannot always get the empty set as a result of an intersection. Axioms in dependency of the case? <brrrrr>
And even in our case, one had to explain why $\{\emptyset\, , \,\mathbb{R}\}\in \tau$ is covered by the other two axioms, so it wouldn't even save lines.

I like to consider $\{\emptyset\, , \,X\}\in \tau$ as the fundamental property, not a resulting one which in general it is not. And it is too important to hide it somewhere.

 

[PeroK]

I like to consider $\{\emptyset\, , \,X\}\in \tau$ as the fundamental property, not a resulting one which in general it is not. And it is too important to hide it somewhere.

Ugly or not, I've seen it done.

 

[PeroK]

E.g. here:

https://math.stackexchange.com/questions/3436996/minimal-axioms-for-topology-open-sets-definition

 

[fresh_42]

E.g. here:

https://math.stackexchange.com/questions/3436996/minimal-axioms-for-topology-open-sets-definition

$\emptyset \subseteq \tau,$ really? And $X:=\cup \tau$ is more artificial than $X\in \tau$ is. It does not work without that setting, since $\left(\left\{a,b\right\};\left\{\emptyset,\{a\},\{a,b\}\right\}\right)$ is a topological space and $\left(\left\{a,b\right\};\left\{\{a\},\{a,b\}\right\}\right)$ or $\left(\left\{a,b\right\};\left\{\emptyset,\{a\}\right\}\right)$ are not.

This is not minimal, these are hidden assumptions, shell game tricks.

 

[WWGD]

My point, for what it's worth, is that you can drop the first axiom about the empty set and full set being in the topology - as long as you recognise the empty intersection and empty union in the other axioms.

It doesn't practically change what you have to test, since checking the empty intersection and empty union are special cases that amount to checking that the empty set and full set are in the topology! But, you can save yourself an axiom if you want.

The full set will necessarily result from the union of all sets in the topology. The empty set then can be derived through DeMorgan or similar.

 

[fresh_42]

The full set will necessarily result from the union of all sets in the topology. The empty set then can be derived through DeMorgan or similar.

This is a tautology. "The full sets results from the topology." You need to define $X:=\cup \tau.$ Otherwise, you simply do not get the full set. There is no way to get $\{a,b\}$ from $\tau\stackrel{?}{=}\{\emptyset,\{a\}\}$ without putting it into $\tau$ first.

Every other view on it is shell game mathematics.

 

[WWGD]

This is a tautology. "The full sets results from the topology." You need to define $X:=\cup \tau.$ Otherwise, you simply do not get the full set. There is no way to get $\{a,b\}$ from $\tau\stackrel{?}{=}\{\emptyset,\{a\}\}$ without putting it into $\tau$ first.

Every other view on it is shell game mathematics.

You don't need to define it that way. Once you're provided with all sets in the topology, the definition warrants that their union, which will include the entire space, is in the topology.

 

[fresh_42]

You don't need to define it that way. Once you're provided with all sets in the topology, the definition warrants that their union, which will include the entire space, is in the topology.

No. You use a tautology to achieve that result. What is wrong with my example?

Space $X:=\{a,b\}.$ Open sets $U_1=\emptyset$ and $U_2=\{a\}.$

Now construct a topology from that without defining $U_3=:X.$

Shell game, but neither logic nor mathematics.

 

[WWGD]

No. You use a tautology to achieve that result. What is wrong with my example?

Space $X:=\{a,b\}.$ Open sets $U_1=\emptyset$ and $U_2=\{a\}.$

Now construct a topology from that without defining $U_3=:X.$

Shell game, but neither logic nor mathematics.

Every element in the space has to be " covered " ( included in) some open set. Otherwise missing elements are somehow extraneous. If you can't define neighborhoods of some points, as in b here, those points are not part of your topological space. But this all ultimately becomes more of an issue of semantics, maybe philosophy. Not sure how productive these exchanges can be.

 

[PeroK]

... well, it's actually the intersection of no sets that produces the full set, but that point seems to have been lost in the general consternation.

 

[PeroK]

... and the union of no sets produces the empty set.

 

[fresh_42]

Every element in the space has to be " covered " ( included in) some open set. Otherwise missing elements are somehow extraneous. If you can't define neighborhoods of some points, as in b here, those points are not part of your topological space. But this all ultimately becomes more of an issue of semantics, maybe philosophy. Not sure how productive these exchanges can be.

Shell game. My speech. "Put them into have them in." A ridiculous argument in my mind. Fact is: you cannot generate $\{a,b\}$ without defining that $b$ has to be part of some open set? And how is that any different from $\{a,b\}\in \tau?$

 

[PeroK]

... although the above link highlights some issues with the former on set-theoretic grounds.

 

[WWGD]

Shell game. My speech. "Put them into have them in." A ridiculous argument. Fact is: you cannot generate $\{a,b\}$ without defining that $b$ has to be part of some open set? And how is that any different from $\{a,b\}\in \tau?$

Its more ridiculous to define a topology that does not allow you to include elements to which it applies. How about defining your topology on the Reals themselves with just the empty set and $\{a\}$? Then you can't really speak of anything but the element a itself. Topology on X is a scheme to say something about X and its elements. But if you don't include an element in an open set, you can't say anything about it, which is just absurd. I guess to me you start with a set and define a topology in it, not the other way around.

 

[fresh_42]

Its more ridiculous to define a topology that does not allow you to include elements to which it applies.

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!

 

[WWGD]

But this discussion doesn't in my view lead to anything productive, so I'm bowing out, at least for now. You have your views, I have mine, and I don't see us making any headway here.

 

[Orodruin]

Its more ridiculous to define a topology that does not allow you to include elements to which it applies. How about defining your topology on the Reals themselves with just the empty set and $\{a\}$? Then you can't really speak of anything but the element a itself. Topology on X is a scheme to say something about X and its elements. But if you don't include an element in an open set, you can't say anything about it, which is just absurd. I guess to me you start with a set and define a topology in it, not the other way around.

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.

 

[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.