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I want to compute the dimension of $\mathbb{A}_{\mathbb{C}}^{1}$, that is the dimension of the affine space in 1 dimension over the field $\mathbb{C}$ but with respect the $\textbf{Euclidean}$ topology.

My definition of dimension of a topological space $X$: is the greatest upper bound of the set of all ascending chains of length n that consists of irreducible non-empty and closed subsets of X ( $C_{0} \subset C_{1} \ .... \subset C_{n}$).

Let $A \subset \mathbb{C}$ with say two points, then $A = \{x,y\} = \{x\} \cup \{y\}$. But $\mathbb{C}$ is a metric space hence every singleton is closed. Thus if A has at least two points then A is reducible. So the only irreducible sets are singletons because $\mathbb{C}$ with the Euclidean topology is not irreducible. Therefore the dimension is equal to $0$.

Is the above correct? Also, another question, in general is there a quick or easy way to figure out the dimension of a given topological space?


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up vote 6 down vote accepted

The notion of dimension of a topological space you are dealing with is not the one (or any of the ones) that topologists usually deal with. For instance, for most reasonable notions of topological dimension (e.g. covering dimension) the Euclidean space $\mathbb{R}^n$ has dimension $n$ (and thus $\mathbb{C}^n \cong \mathbb{R}^{2n}$ has dimension $2n$). Let us call the invariant you have defined the Krull dimension of a topological space.

Edit: in this answer I want to make reference to two different topologies on $\mathbb{C}$. It seems best to use separate notation for them. Let me define

$\mathbb{C}_E$ to be $\mathbb{C}$ endowed with its standard (Euclidean) topology, and
$\mathbb{C}_Z$ to be $\mathbb{C}$ endowed with the Zariski topology, or equivalently the cofinite topology: the proper closed sets are precisely the finite sets.

There are a couple of problems with your computation of the Krull dimension of $\mathbb{C}_E$. First there is an issue with the indexing: you seem to be claiming that the only nonempty irreducible closed subsets of $\mathbb{C}_E$ are the singleton sets. But if this is true, this means the maximal length of a chain is $0$, not $1$, and thus the Krull dimension of $\mathbb{C}_E$ would be zero.

Second: your claim is true, but your argument for it does not seem correct to me. What you have said is that in a separated space -- i.e., a space in which all singletons are closed (others say "$T_1$") -- no $2$-element subset is irreducible. That's true, but there could be larger irreducible closed subsets. Indeed, the space $\mathbb{C}_Z$ is separated and irreducible: it cannot be the union of two proper closed subsets, since it is infinite and every proper closed subset is finite. Your argument works equally well to show that a finite closed set in a separated space is only irreducible if it consists of a single point, so the maximal chains of irreducible closed subsets in $\mathbb{C}_Z$ are $\{x\} \subset \mathbb{C}$ and thus the Krull dimension is $1$. (More generally, if you endow $\mathbb{C}^n$ with the Zariski topology, it has Krull dimension $n$.)

So separatedness does not distinguish between $\mathbb{C}_E$ and $\mathbb{C}_Z$. You need to use some stronger property of metric topologies here, and the next most obvious guess turns out to be correct.

Exercise: If $X$ is a Hausdorff space, then the only irreducible closed subsets are the singleton sets.

You should try to prove this on your own and ask if you have any difficulties.

The exercise implies that the Krull dimension of any Hausdorff space is zero.

For a little more information on irreducible spaces, see $\S 13.4$ of these notes.

As for your last question: well, most spaces one meets in conventional topology are Hausdorff, so have Krull dimension zero. Probably the most important class of spaces which need not be Hausdorff (especially for an algebraic geometer!) are the spaces of the form $\operatorname{Spec} R$ for $R$ a commutative ring (i.e., Zariski topologies). Here the name "Krull dimension" is justified by the fact that the Krull dimension of $\operatorname{Spec} R$ is equal to the Krull dimension of $R$ in the sense of ring theory, i.e., computed in terms of lengths of chains of prime ideals. This follows from the correspondence between irreducible closed subsets and prime ideals. Finally, if you have a space $X$ which is a finite union of open subspaces $X_1,\ldots,X_n$, then the Krull dimension of $X$ is equal to the maximum of the Krull dimensions of the $X_i$'s. This helps to compute the Krull dimension of the underlying topological space of an algebraic variety or a sufficiently reasonable scheme. In practice, if you are given a scheme then its Krull dimension is one of the easiest things to compute: if you don't know that, then you really don't know much about your scheme.

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Thanks! If $X$ is a Hausdorff space, then the only irreducible closed subsets are the singleton sets. Suppose $A \subset X$ is irreducible with at least two points. Say $x,y \in A$. Since $X$ is Hausdorff we can find disjoint open sets $U_{x}$ and $V_{y}$ containing $x$ and $y$ respectively. Now $U_{x} \cap A$ and $V_{y} \cap A$ are non-empty open sets in $A$. But $A$ is irreducible so these open sets intersect, but this is absurd because they are disjoint. I used the fact that a space $X$ is irreducible iff every two non-empty sets have non-empty intersection. Is this OK? – user6495 Feb 19 '11 at 21:24
I meant two non-empty open sets have non-empty intersection. – user6495 Feb 19 '11 at 21:46
@user6495: sounds good to me. – Pete L. Clark Feb 19 '11 at 22:20

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