Subspace Top (understanding a problem in the book)

I think I understand this example from the book. I just need to verify.

Let . The dictionary order on is just the restriction to of the dictionary order on the plance . However, the dictionary order topology is not the same as the subspace topology on obtained from the dictionary order on . For example, the set is open in in the subspace, but not in the order topology.

Ok so the subspace is of the form where

Take .

Now, the intersection is , but why is this set considered open in the subspace topology when it is have closed?

However, the order topology only has half closed or open sets if the member is the least or largest element respectively. The largest element would be . Therefore, the set isn't open in the order topology.

Re: Subspace Top (understanding a problem in the book)

Quote:

Originally Posted by

**dwsmith** I think I understand this example from the book. I just need to verify.

Let

. The dictionary order on

is just the restriction to

of the dictionary order on the plance

. However, the dictionary order topology is not the same as the subspace topology on

obtained from the dictionary order on

. For example, the set

is open in

in the subspace, but not in the order topology.

Ok so the subspace is of the form

where

Take

.

Now, the intersection is

, but why is this set considered open in the subspace topology when it is have closed?

However, the order topology only has half closed or open sets if the member is the least or largest element respectively. The largest element would be

. Therefore, the set isn't open in the order topology.

So, I don't quite know what as you wrote it is, but if you let then by definition (being just an interval) one has that is open in with the dictionary ordering, and since we have that our set in question is the intersection of an open set in with and so, by definition, open in the subspace topology. I also am not quite sure what your last paragraph means, please try to reexplain it--I believe it's what I'm about to say though. To me what makes the most obvious sense is that if we suppose that to be our set in question, then if were open in with the dictionary ordering we can find some basic open set with . But, since is not a maximal member of we know that we can take to be an open interval, and so in particular there exists with which contradicts . More visually appealing, any open set containing spills onto the next vertical line in .

Re: Subspace Top (understanding a problem in the book)

Quote:

Originally Posted by

**Drexel28** So, I don't quite know what

as you wrote it is, but if you let

then by definition (being just an interval) one has that

is open in

with the dictionary ordering, and since

we have that our set in question is the intersection of an open set in

with

and so, by definition, open in the subspace topology. I also am not quite sure what your last paragraph means, please try to reexplain it--I believe it's what I'm about to say though. To me what makes the most obvious sense is that if we suppose that

to be our set in question, then if

were open in

with the dictionary ordering we can find some basic open set

with

. But, since

is not a maximal member of

we know that we can take

to be an open interval, and so in particular there exists

with

which contradicts

. More visually appealing, any open set containing

spills onto the next vertical line in

.

Is x an interval?

Re: Subspace Top (understanding a problem in the book)

Quote:

Originally Posted by

**dwsmith** Is x an interval?

No, just an element of .

Re: Subspace Top (understanding a problem in the book)

Quote:

Originally Posted by

**dwsmith** I think I understand this example from the book. I just need to verify.

Let

. The dictionary order on

is just the restriction to

of the dictionary order on the plance

. However, the dictionary order topology is not the same as the subspace topology on

obtained from the dictionary order on

. For example, the set

is open in

in the subspace, but not in the order topology.

Ok so the subspace is of the form

where

Take

.

Now, the intersection is

, but why is this set considered open in the subspace topology when it is have closed?

However, the order topology only has half closed or open sets if the member is the least or largest element respectively. The largest element would be

. Therefore, the set isn't open in the order topology.

Consider the point . What would an open ball look like in and what would it look like in around that point? In the former, there are no points greater than 1, so even though it looks like you are in a half-closed interval, there are no points greater than it to include in your open ball. Therefore, any epsilon you choose small enough gives you only points in your set. On the other hand, in , any open ball at that same point has an infinite number of points that are not in your set, so it is not open there.