# Dimension of a topological space

1. Feb 11, 2008

### matness

In Hartshorne's book definiton of a dimension is given as follows:
İf X is a t.s. , dim(X) is the supremum of the integers n s.t. there exist a chain
$$Z_0 \subsetneq Z_1.....\subsetneq Z_n$$
of distinct irreducible closed subsets of X

My question is:
Can we conclude directly that any topological space has dim greater than or equal to 1, since empty set and and X itself is always closed?

Example in the same book says no in a way.It says A^1 has dim 1.
Although $$\emptyset \subsetneq {any point} \subsetneq X$$
Should I exclude empty set ?

2. Feb 11, 2008

### HallsofIvy

Staff Emeritus
What is your definition of "irreducible closed subset"?

3. Feb 11, 2008

### matness

yes.that is the answer.i was careless as always.
another question then: can we say empty set is reducible then. since it is not irreducible
Or do we exclude this set from the discussion of reducibility/irreducibility?

(my guess is second choice is less problematic)

4. Feb 18, 2008

### eastside00_99

This is one definition for dimension of algebraic varieties. There are several others...such as the Krull Dimension of the variety's associated coordinate ring.

The empty set an irreducible algebraic variety. Irreducibility of an algebraic variety V means that if V=W_1 union W_2, then W_1=empty set or W_2=empty set.

I believe that the definition of dimension that you provide does generalize to topology but as you have already mentioned you have to determine what it means to be irreducible. But, yeah you either have to throw out the overall space or the empty set.

5. Feb 20, 2008

### HallsofIvy

Staff Emeritus
You still haven't answered my question: what is the definition of irreducible closed subset? And, since you mention it, what is the definition reducible closed subset?

6. Feb 20, 2008

### eastside00_99

Irreducibility is a property of algebraic varieties. I will post what I put in my previous post again:

Irreducibility of an algebraic variety V means that if V=W_1 union W_2, then W_1=empty set or W_2=empty set.

There may be other equivalent ways of saying it.

The way to talk about the situation for topological spaces is much more complicated. You need something analogous to closed balls in R^n. I.e., closed sets whose boundary has dimension 1 less than the closed set. Then the next closed ball is going to be contained in the other boundary.

7. Feb 20, 2008

### Hurkyl

Staff Emeritus
Hartshorne's definition of "irreducibile" and "dimension" are for arbitrary topoplogical spces. matness stated the definition of "dimension" correctly. (He has not yet stated Hartshorne's definition of "irreducible")

Yes, there are several other ways to define the notion of "dimension" for topological spaces -- but those other ways are not the ones Hartshorne is using in his text.

Last edited: Feb 20, 2008
8. Feb 20, 2008

### eastside00_99

ah, I see:

irreducibility for a topological space must then be taking the definition I provided above except changing the words "algebraic set" to be closed set.

9. Feb 20, 2008

### Hurkyl

Staff Emeritus
Effectively; the exception is that, in Hartshorne, the empty set is defined to be not irreducible.

For (traditional) varieties, of course, they're the same thing: the closed sets in the Zariski topology are precisely the algebraic sets. Much of Hartshorne works with schemes (a generalization of varieties) but he states these topological definitions in full generality (and they might be relevant in the sections where he talks about ringed spaces).

10. Feb 20, 2008

### eastside00_99

Well, I don't see the problem with letting the empty set be considered a scheme. We trivially have a sheaf of modules associate with {} and we trivially have the empty topology ({},{{}}). But, yeah, if you are going to work with schemes using topological definitions makes sense.

11. Feb 23, 2008

### matness

A 'nonempty' subset of a topological space is irreducible if it can not be written as union of its two proper closed subsets.

Because of the word 'nonempty' the argument in my first post is useless. And while writing second post i took definition of reducible as not being irreducible

12. Feb 23, 2008

### matness

Although I can not see directly( or indirectly) , they should be equivalent to be consistent in topology, arent they?

13. Feb 23, 2008

### Hurkyl

Staff Emeritus
There's nothing wrong with having several different notions of dimension.

Probably the most obvious example of this is that sometimes we want to consider C as a two-dimensional space, and sometimes as a one-dimensional space.

(Hartshorne's definition is usually only useful for Noetherian spaces -- according to it, C is infinite dimensional. But C with the Zariski topology is 1-dimensional)

14. Jul 10, 2008

### Take_it_Easy

I just discovered this thread where the topological dimension is playing a role. I am interested in this question:

Let $${\mathbb I} = {\mathbb R} \setminus {\mathbb Q}$$ the set of the irrational numbers of the real line.

What is the topological dimension of
$${\mathbb R}^2 \setminus {\mathbb I} \times {\mathbb I}$$ ????

15. Jul 10, 2008

### mathwonk

you just have to take the right notion of subspaces. dimension of an algebraic variety is the length of a maximal nested seqeunce of irreducible algebraic subvarieties.

the dimension of a linear space is the length of a maximal sequence of linear subspaces.

so the (linear) dimension of C is the maximal length of a sequence of linear subspaces of C, hence the real linear dimension is 2 and the complex linear dimension is 1.