1. Sep 23, 2006

### ak416

Ok, I was just wondering. They say in my book to think of a strictly finer topology of T to be the same as T but some of the sets in T are divided into more peices. But can't you have a strictly finer topology of T that consists of the same elements as T but the finer topology has 1 set that properly contains a set in T (Its only intersection would be the sets it properly contains and of course X, the topological space). Also, more than 1 set with this property should also work to make it a stricly finer topology than T. So if this works, then can't you think of a stricly finer topology of T as being one that contains T but also has "bigger sets" instead of the "finer sets"?

Note: I added an example in the picture attachment. All sets in black are T (plus their unions and intersections). The blue set is a set of the finer topology.

#### Attached Files:

• ###### top1.GIF
File size:
1.6 KB
Views:
29
Last edited: Sep 23, 2006
2. Sep 23, 2006

### StatusX

I'm not sure what you're asking. When you add sets to a topology on a space, you add both smaller and larger sets, since the new space must be closed under unions and finite intersections. For example, adding all points of R to the usual topology on R (ie, making points open sets) gives the discrete topology, where every subset of R is open. This is strictly finer than the usual topology, and contains both smaller sets (like points) and larger ones (like [0,1]).

3. Sep 23, 2006

### ak416

Ok, imagine T is all the black sets plus their unions and intersections. Now add the blue set. Is T' (all the black sets plus the blue set and all arbitrary unions and intersections of these sets) Strictly finer than T? That's my question.

Last edited: Sep 23, 2006
4. Sep 23, 2006

### mr_coffee

ak,

So you don't have to wait for the images to be approved, you might want to try to just post them here:
http://imageshack.us/
or
http://www.suprfile.com/

THen just copy and paste the link into the insert image button.

5. Sep 23, 2006

### ak416

6. Sep 23, 2006

### StatusX

Of course it is. It contains all the open sets of the old topology plus some more, which is all it means for a topology to be strictly finer.

The way I understood your original question was like this: If T1 and T2 are two topologies on X, and if T2 is strictly finer than T1, is there an open set A in T1 such that a proper subset of A is open in T2 but not in T1? Is that right? Well in a trivial sense this is true, since the sets you add are proper subsets of X that are open in T2 but not in T1. So you are subdividing the open set X into smaller pieces. But none of the proper subsets of X get subdivided in your example, so in that sense, the statement is false.

7. Sep 23, 2006

### ak416

ok i thought so as well. Its just the way they describe finer topologies in the book is they say imagine a bunch of rocks (and their unions) to be the topology T. Then the finer topology T' would be finer in the sense that the rocks are grinder into smaller pebbles. But the topology I showed you here is not like that. The finer one actually contains bigger "rocks" instead of the original ones divided into smaller pieces. So i just wanted to double check my interpretation...

8. Sep 23, 2006

### matt grime

But you're analogy fails you because you have not considered what the 'finest' rocks are, that your definition generates.

If you just take the standard metric topology on R, and then add in the rocks {X: X is the closure of some open set in the metric topology}, and then take the toplogy it generates, you get the discrete topology, even though you've only described 'bigger' rocks. You have failed to think about the totality of open sets in the topology, and have instead cherry picked some subset of them.