Dimension

edit_note

This post is being revised. The text below is the version as of 2026-06-10, so some statements may be out of date and references to this post from other posts may not line up.

This post was machine-translated from the Korean original by Marvin (via Kimi). It may contain errors or awkward phrasing — the Korean original is the source of truth.

In this post we define the dimension of a topological space. First we define the dimension commonly used in general topology, and then we separately define the notion of dimension used in Algebraic Varieties.

Covering dimension

For convenience, following [Mun] we define the dimension only for compact spaces in this section. The basic idea is to measure how many times a point of \(X\) is covered by open sets; of course \(X\) is covered by the single open set \(X\) itself, so we must define this using arbitrary open coverings, and if we allow arbitrary open coverings then a single point can be covered by as many open sets as we wish, so we need to guarantee some kind of minimality.

First we define the following.

Definition 1 A family \((U_i)_{i\in I}\) of subsets of \(X\) is said to have order \(m+1\) if no point of \(X\) lies in more than \(m+1\) of the \(U_i\), and some point of \(X\) lies in exactly \(m+1\) of the \(U_i\).

Then we can define the dimension of a space \(X\) as follows.

Definition 2 A space \(X\) is finite dimensional if there exists some \(m\) such that, whenever an open covering \((U_i)_{i\in I}\) of \(X\) is given, there always exists an open refinement \((V_j)_{j\in J}\) of \((U_i)\) of order \(m+1\). We define the smallest such \(m\) to be the dimension of \(X\), and write \(\dim X\).

Something worth noting is that topological spaces can look quite strange, as in the following figure, and in such cases we define the dimension of the space to be the larger of the dimensions of the two components.

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On the other hand, there are several properties we expect dimension to satisfy, some of which are as follows.

Proposition 3 If \(X\) is a finite dimensional topological space and \(Y\) is a closed subspace of \(X\), then \(Y\) is also finite dimensional and \(\dim Y\leq\dim X\).

Proof

Suppose \(X\) is a \(d\)-dimensional topological space, and let \(\{V_j\}\) be an arbitrary open covering of \(Y\). Then for each \(V_j\) there exists an open subset \(U_j\) of \(X\) such that \(V_j=U_j\cap Y\). Now \(X\) can be covered by the \(U_j\) together with \(X\setminus Y\). Then there exists a refinement of this covering of order \(\leq d+1\), and intersecting this again with \(Y\) gives a refinement of \(\{V_j\}\) of order \(\leq d+1\).

The following proposition is a more mathematically polished version of the caution mentioned after Definition 2.

Proposition 4 If there exist two finite dimensional closed subspaces \(Y,Z\) of a topological space \(X\) such that \(X=Y\cup Z\), then \(\dim X=\max(\dim Y,\dim Z)\).

Proof

And of course we hope that the dimension of \(\mathbb{R}^n\) is \(n\). However, showing this is not easy, essentially because it is already difficult for us to show that \(\mathbb{R}^n\) and \(\mathbb{R}^m\) are not homeomorphic. Instead, the following weaker proposition can be easily seen from the definition.

Proposition 5 Any compact subspace of \(\mathbb{R}^n\) always has dimension at most \(n\).

Proof

Krull dimension

On the other hand, we will define the notion of dimension used in algebraic geometry; the spaces of interest in algebraic geometry are generally equipped with a topology different from the usual one, so this definition is somewhat non-intuitive. In particular, \(\mathbb{R}^n\) with its usual topology is always \(0\)-dimensional. Nevertheless, it is a fact that this definition can be phrased in the language of topology, so we record it here as well.

Definition 6 A topological space \(X\) is irreducible if there do not exist nontrivial closed subsets \(A,B\) of \(X\) such that \(X=A\cup B\).

Then the following are all equivalent.

Proposition 7 For a topological space \(X\), the following are all equivalent.

1. \(X\) is irreducible.
2. For any nonempty open subsets \(U,V\) of \(X\), we have \(U\cap V\neq\emptyset\).
3. For any nonempty open subset \(U\) of \(X\), we have \(\cl U=X\).
4. Every open subset of \(X\) is connected.

Proof

The equivalence of the first and second conditions is obvious by considering complements, and the equivalence of the second and third conditions is obvious by considering \(X\setminus \cl U\) and \(U\). Finally, the equivalence of the second and fourth conditions follows from the definition.

In particular, an irreducible space is not Hausdorff. Because of the last equivalence in the above proposition, irreducible spaces are also called hyperconnected spaces. In a similar vein, the following holds. (See: §Connected Spaces, ⁋Proposition 3)

Proposition 8 Suppose \(X\) is a union of irreducible open subsets

\[X=\bigcup_{i\in I} U_i\]

and \(U_i\cap U_j\neq \emptyset\) holds for all \(i,j\). Then \(X\) is irreducible.

Proof

Let \(V, W\) be arbitrary open sets; we show that \(V\cap W\neq\emptyset\). Then by the given assumption there first exist \(i,j\) satisfying \(U_i\cap V\neq\emptyset\) and \(U_j\cap W\neq\emptyset\). Now by the third result of Proposition 7 and §Subspaces, \(U_i\) is also irreducible, so the two nonempty subsets \(U_i\cap V\) and \(U_i\cap U_j\) of \(U_i\) must have a nonempty intersection. That is,

\[(U_i\cap V)\cap (U_i\cap U_j)=U_i\cap U_j\cap V\neq\emptyset\]

and viewing \(U_i\cap U_j\cap V\) as a nonempty subset of \(U_j\), similarly from the irreducibility of \(U_j\) we obtain the equality

\[(U_i\cap U_j\cap V)\cap (U_j\cap W)=U_i\cap U_j\cap V\cap W\neq\emptyset\]

and in particular \(V\cap W\neq\emptyset\).

Similarly to connected components, we can define the following.

Definition 9 For a subset \(A\) of a topological space \(X\), the irreducible component containing \(A\) means the largest irreducible subset containing \(A\).

Then by an argument similar to §Connected Spaces, ⁋Proposition 2, we can show that the closure of an irreducible set is irreducible, so an irreducible component must be a closed subset.

Definition 10 For a topological space \(X\), we define the length of a strictly descending chain of irreducible closed subsets of \(X\)

\[A_n\supsetneq\cdots\supsetneq A_0\]

to be \(n\). Then we define the Krull dimension of \(X\) by the formula

\[\dim X=\sup\{\text{length of strictly descending chains of irreducible closed subsets}\}\]

If there exists a strictly descending chain of infinite length, we define \(\dim X=\infty\), and if \(X=\emptyset\) we define \(\dim X=-\infty\).

Then in the following situation the Krull dimension of \(X\) is always finite. In particular, in a Hausdorff space only singletons are irreducible subsets, so the Krull dimension of a Hausdorff space is \(0\).

Definition 11 A topological space \(X\) is noetherian if whenever a chain of closed subsets

\[A_1\supseteq A_2\supseteq\cdots\]

is given, there exists some \(n\) such that \(A_n=A_{n+1}=\cdots\).

The noetherian condition gives a strong finiteness condition. For example, the following holds.

Proposition 12 A noetherian space is compact.

Proof

Suppose a noetherian space \(X\) and an open covering \(\{U_i\}_{i\in I}\) of \(X\) are given. Then we can define

\[\mathcal{C}=\left\{\bigcup_{j\in J} U_j:\text{$J$ finite subset of $I$}\right\}\]

Now considering an arbitrary totally ordered subset of \(\mathcal{C}\), this is equivalent to a descending chain of closed sets given by their complements, and therefore by the assumption that \(X\) is noetherian this must eventually stop. That is, \(\mathcal{C}\) satisfies the condition of [Set Theory] §Axiom of Choice, ⁋Theorem 4 (Zorn’s lemma) and therefore \(\mathcal{C}\) has a maximal element \(U\in \mathcal{C}\). If \(X\neq U\), then we can choose some \(U_j\) containing \(x\in X\setminus U\), and then \(U\cap U_j\) is an element of \(\mathcal{C}\) that strictly contains \(U\), contradicting the maximality of \(U\). Therefore \(U=X\) and we obtain the desired result.

Additionally, the following holds for noetherian spaces.

Proposition 13 For a noetherian topological space \(X\), the following hold.

1. Any subspace of \(X\) is noetherian.
2. \(X\) has finitely many irreducible components.
3. Each irreducible component of \(X\) contains a nonempty open subset of \(X\).

Proof

1. Let \(Y\) be an arbitrary subspace of \(X\) and suppose a descending chain of closed subsets of \(Y\)

   \[A_1\supseteq A_2\supseteq \cdots\]

   is given. Then there exist closed subsets \(A_i'\) of \(X\) satisfying \(A_i=A_i' \cap Y\). Now setting \(B_i=A_1'\cap\cdots\cap A_i'\), we have \(B_i\cap Y=A_i\) and the \(B_i\) form a descending chain of closed subsets of \(X\).

2. Let \(\mathcal{C}\) be the collection of closed subsets of \(X\) that cannot be expressed as a finite union of irreducible components. Then we must show that \(\mathcal{C}=\emptyset\). Suppose for contradiction that \(\mathcal{C}\) is nonempty; then by the same method as in the proof of Proposition 12 we know that \(\mathcal{C}\) has a minimal element \(A\). Now \(A\) is not irreducible, so we can write \(A=B_1\cup B_2\) for two closed subsets \(B_1,B_2\), and by the assumption that \(B_1,B_2\not\in \mathcal{C}\) each of these has finitely many irreducible components. Through a small computation we can use these irreducible decompositions to find a finite irreducible decomposition of \(A=B_1\cup B_2\), which is a contradiction, so we must have \(\mathcal{C}=\emptyset\).
3. Suppose \(X=A_1\cup\cdots\cup A_n\) is an irreducible decomposition; then considering \(X\setminus (A_2\cup\cdots\cup A_n)\), this is a nonempty open subset of \(X\) contained in \(A_1\).

Then if \(X\) is noetherian, there exists an irreducible decomposition

\[X=\bigcup_{i=1}^r X_i\]

where the \(X_i\) are all closed, and since the complement of each \(X_i\) is also a finite union of closed sets, each \(X_i\) is also open.

Proposition 14 For a topological space \(X\) and an open subset \(U\), there exists a bijection between irreducible closed subsets of \(X\) meeting \(U\) and irreducible closed subsets of \(U\).

Proof

First, suppose an irreducible subspace \(Z\) of \(X\) satisfying \(U\cap Z\neq\emptyset\) is given; we must show that any two nonempty open subsets of \(Z\cap U\) are not disjoint. Any subset of \(Z\cap U\) is of the form \(V_1\cap U\), \(V_2\cap U\) for open subsets \(V_1, V_2\) of \(Z\), so from the equality

\[(V_1\cap U)\cap (V_2\cap U)=(V_1\cap V_2)\cap U\]

if \((V_1\cap U)\cap(V_2\cap U)\neq\emptyset\) then \(V_1\cap V_2\neq\emptyset\), contradicting the assumption that \(Z\) is irreducible.

Conversely, suppose an irreducible closed subset \(Y\subseteq U\) of \(U\) is given; then the closure of \(Y\) is also irreducible, so \(\cl_X(Y)\) becomes an irreducible subset of \(X\) meeting \(U\). That is, from this we obtain two functions

\[\{\text{irreducible closed subset of $X$ meeting $U$}\}\rightarrow \{\text{irreducible closed subset of $U$}\};\qquad Z\mapsto Z\cap U\]

and

\[\{\text{irreducible closed subset of $U$}\} \rightarrow \{\text{irreducible closed subset of $X$ meeting $U$}\};\qquad Y\mapsto \cl_X(Y)\]

and we can verify that these are bijections inverse to each other.

Moreover, since the two functions in the above proof preserve inclusion, there exists a bijection between irreducible components of \(X\) meeting \(U\) and irreducible components of \(U\).

Now let us show the following proposition.

Proposition 15 If there exist two finite dimensional closed subspaces \(Y,Z\) of a topological space \(X\) such that \(X=Y\cup Z\), then their Krull dimensions also satisfy the formula \(\dim X=\max(\dim Y,\dim Z)\).