Standard topology is coarser than lower limit topology?

[patric44]

Homework Statement

Standard topology is coarser than lower limit topology?

Relevant Equations

T={u subset R: for all x in u exists d>0 s.t. (x-d,x+d) subset u}

Hello everyone,
Our topology professor have introduced the standard topology of $\mathbb{R}$ as:
$$\tau=\left\{u\subset\mathbb{R}:\forall x\in u\exists\delta>0\ s.t.\ \left(x-\delta,x+\delta\right)\subset u\right\},$$
and the lower limit topology as:
$$\tau_{1}=\left\{u\subset\mathbb{R}:\forall x\in u\exists\delta>0\ s.t.\ \left[x,x+\delta\right)\subset u\right\}.$$
He asked for the relation between the two topologies. It is easy to show that $\tau_{1}$ is coarser than $\tau$ according to this definition:
$$\left(x-\delta,x+\delta\right)\subset \left[x,x+\delta\right)\subset v,$$
for $v$ in $\tau_{1}$. But that is not true in all topology references that I read. According to their definition (the collection of all open intervals in the real line forms a standard topology) the standard topology is coarser than lower limit topology.
Will appreciate any help, thanks.

 

[fresh_42]

Homework Statement:: Standard topology is coarser than lower limit topology?
Relevant Equations:: T={u subset R: for all x in u exists d>0 s.t. (x-d,x+d) subset u}

Hello everyone,
Our topology professor have introduced the standard topology of $\mathbb{R}$ as
$$\tau=\left\{u\subset\mathbb{R}:\forall x\in u\exists\delta>0\ s.t.\ \left(x-\delta,x+\delta\right)\subset u\right\},$$
and the lower limit topology as
$$\tau_{1}=\left\{u\subset\mathbb{R}:\forall x\in u\exists\delta>0\ s.t.\ \left[x,x+\delta\right)\subset u\right\}.$$
He asked for the relation between the two topologies. It is easy to show that $\tau_{1}$ is coarser than $\tau$ according to this definition:
$$\left(x-\delta,x+\delta\right)\subset \left[x,x+\delta\right)\subset v$$

$$\left(x-\delta,x+\delta\right)\supset \left[x,x+\delta\right)$$

for $v$ in $\tau_{1}$. But that is not true in all topology references that I read. According to their definition (the collection of all open intervals in the real line forms a standard topology) the standard topology is coarser than lower limit topology.
Will appreciate any help, thanks.

 

[patric44]

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.

 

[fresh_42]

Every open set in the standard topology $\tau$ is an open set in the lower limit topology $\tau_1,$ i.e. $\tau \subseteq \tau_1,$ i.e. $\tau$ is coarser than $\tau_1.$

 

[patric44]

Every open set in the standard topology $\tau$ is an open set in the lower limit topology $\tau_1,$ i.e. $\tau \subseteq \tau_1,$ i.e. $\tau$ is coarser than $\tau_1.$

Actually I'm looking for a proof for this statement according to the given definition.

According to their definition (the collection of all open intervals in the real line forms a standard topology) the standard topology is coarser than lower limit topology.

 

[fresh_42]

Let $x\in U\in \tau.$ Then $[x,x+\delta)\subseteq (x-\delta,x+\delta) \subseteq U$ for some $\delta > 0.$ However, this is the definition of an open set in $\tau_1.$ Hence $\tau \subseteq \tau_1$ and $|\tau| \leq |\tau_1|,$ which means $\tau$ is coarser as it has fewer open sets.

 

[FactChecker]

Actually I'm looking for a proof for this statement according to the given definition.

Start with an arbitrary open interval $(x-\gamma, x+\gamma) \in \tau$. Using the fact that the union of infinitely many open sets is open, can you show that this interval is open in the lower limit topology?

 

[fresh_42]

Start with an arbitrary open interval $(x-\gamma, x+\gamma) \in \tau$. Using the fact that the union of infinitely many open sets is open, can you show that this interval is open in the lower limit topology?

This isn't even necessary here as we have given an explicit description of all open sets in both topologies.

 

[PeroK]

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.

I suspect you may have misunderstood something quite fundamental about the notation employed here.

 

[PeroK]

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.

PS if we take $x = 0$, then the sets are $[0, 1)$ and $(-1, 1)$ and if you don't see that the former is a subset of the latter, then you have not understood the notation.

 

[patric44]

Thanks all, it's clear now, I really appreciate this. The confusion was about the delta notation.

 

[fresh_42]

No. We have a definition of a basis for each topology. It is now necessary to show that a basis element of the standard topology is open in the lower limit topology.

That was what I first thought, too, but it is wrong. We have a description of any open set, not only basis: read the qualifiers! Open is, what contains an open (half-open) interval of all of its elements. This already covers infinite unions. No basic needed.

 

[FactChecker]

This isn't even necessary here as we have given an explicit description of all open sets in both topologies.

Yes. I stand corrected. Thanks.

 

[PeroK]

No. We have a definition of a basis for each topology. It is now necessary to show that a basis element of the standard topology is open in the lower limit topology.

The original post describes each topology completely. It does not simply provide a basis:

Our topology professor have introduced the standard topology of $\mathbb{R}$ as:
$$\tau=\left\{u\subset\mathbb{R}:\forall x\in u\exists\delta>0\ s.t.\ \left(x-\delta,x+\delta\right)\subset u\right\},$$
and the lower limit topology as:
$$\tau_{1}=\left\{u\subset\mathbb{R}:\forall x\in u\exists\delta>0\ s.t.\ \left[x,x+\delta\right)\subset u\right\}.$$

The definition of an open set in $\tau$ is the standard definition of an open set, not only of an open interval.

 

[fresh_42]

The original post describes each topology completely.

I was first unsure about $\emptyset$ and $\mathbb{R}$ but I think that even these are covered.

 

[PeroK]

I was first unsure about $\emptyset$ and $\mathbb{R}$ but I think that even these are covered.

The requirement for $\emptyset$ and $\mathbb R$ to be in the topology is covered by the empty union and empty intersection respectively. And I thought you were a great fan of vacuous truths!

 

[fresh_42]

The requirement for $\emptyset$ and $\mathbb R$ to be in the topology is covered by the empty union and empty intersection respectively. And I thought you were a great fan of vacuous truths!

They are even listed in $\tau, \tau_1$ without bothering any axioms.

If we write a topology as a set, then they have to be included without intersections and unions, simply because it is a listing. $\mathbb{R}$ is trivially included, and $\emptyset$ by vacuous truth about all elements of $\emptyset.$

Yes, I liked that thought.

 

[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.

RH set is $\{ y: x-1 < y < x+1 \}:$
LH set is $\{ y: x \leq y < x+1 \}$
Can you see it?

Edit: A general point: Some Topologies may not be comparable. Can you think of one such example?

 

[Orodruin]

The requirement for $\emptyset$ and $\mathbb R$ to be in the topology is covered by the empty union and empty intersection respectively. And I thought you were a great fan of vacuous truths!

The point was that they are included in the original description already.

$\emptyset$: For all $x \in \emptyset$ it holds that $[x,x+\delta) \subset \emptyset$ because there are no $x \in \emptyset$.

$\mathbb R$: For all $x \in \mathbb R$ it holds that $[x,x+\delta) \subset \mathbb R$ for some $\delta$ (actually, for all $\delta$).

 

[WWGD]

The point was that they are included in the original description already.

$\emptyset$: For all $x \in \emptyset$ it holds that $[x,x+\delta) \subset \emptyset$ because there are no $x \in \emptyset$.

$\mathbb R$: For all $x \in \mathbb R$ it holds that $[x,x+\delta) \subset \mathbb R$ for some $\delta$ (actually, for all $\delta$).

Or use, for fixed $\delta>0, [x-2\delta, x-\delta) \cap [x,x+\delta)$ is a member of the topology.

 

[WWGD]

Or use, for fixed $\delta>0, [x-2\delta, x-\delta) \cap [x,x+\delta)$ is a member of the topology.

This is a basic property of elements of the topology. They are closed under( finite) intersection.
Edit: A simple search would do:

 

[PeroK]

This is a basic property of elements of the topology. They are closed under( finite) intersection.

Your logic is back to front. You are given a definition of a collection of "open" sets. You have to check that every finite intersection of sets in the collection is in the collection. You cannot assume that. That involves checking that the empty set in the collection.

If you assume that $\tau$ is a topology, then you can assume that $\emptyset \in \tau$ and it makes no sense to prove this using the other assumed properties of the topology.

You made the same mistake I made that @Orodruin pointed out.

 

[WWGD]

Your logic is back to front. You are given a definition of a collection of "open" sets. You have to check that every finite intersection of sets in the collection is in the collection. You cannot assume that. That involves checking that the empty set in the collection.

If you assume that $\tau$ is a topology, then you can assume that $\emptyset \in \tau$ and it makes no sense to prove this using the other assumed properties of the topology.

You made the same mistake I made that @Orodruin pointed out.

Not as I read it. The sets of the given form are stated to be elements of the topology. The question is about the relationship _ between the topologies_ , and not on whether these describe topologies. The intersection I provided just verifies that the empty set is part of the topology.

 

[PeroK]

Not as I read it. The sets of the given form are stated to be elements of the topology. The question is about the relationship _ between the topologies_ , and not on whether these describe topologies. The intersection I provided just verifies that the empty set is part of the topology.

Either you assume $\tau$ is a topology or not. If you do, then $\emptyset$ is assumed to be in the topology without further justification. If not, then you cannot assume the finite intersection property.

 

[WWGD]

Not as I read it. The sets of the given form are stated to be elements of the topology. The question is about the relationship _ between the topologies_ , and not on whether these describe topologies. The intersection I provided just verifies that the empty set is part of the topology.

 

[WWGD]

The initial post does explicitly state these are the open sets in the topology. We are _ not_ being asked to verify these define a topology, but _ we are told_ these are topologies.

 

[WWGD]

" He asked for the relationship between these _Topologies_"

 

[PeroK]

The initial post does explicitly state these are the open sets in the topology. We are _ not_ being asked to verify these define a topology, but _ we are told_ these are topologies.

It was you who posted a proof/justification for $\emptyset \in \tau_1$:

Or use, for fixed $\delta>0, [x-2\delta, x-\delta) \cap [x,x+\delta)$ is a member of the topology.

 

[WWGD]

It was you who posted a proof/justification for $\emptyset \in \tau_1$:

Its a verification. Let's have @Orodruin chime in.

 

[fresh_42]

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.