Complete Varieties

In the Wikipedia page, a complete variety is defined to be a variety $X$ such that for any variety $Y$, the projection map $X \times Y \to Y$ is a closed map. Asking whether this map is closed is related to the natural question of whether the image of a subvariety is still a subvariety. The page also mentions that this is analogous to a characterization of compact topological spaces.

Theorem. Let $X$ be a topological space. Then $X$ is compact if and only if for any topological space $Y$, the projection map $p: X \times Y \to Y$ is a closed map, where $X \times Y$ has the product topology.

Proof. ($\implies$) Assume $X$ is compact, $Z \subseteq X \times Y$, and $y \in Y \backslash p(Z)$. Since $p$ is the projection we know that $X \times \{y\} \subseteq (X \times Y) \backslash p^{-1}(p(Z)) \subseteq (X \times Y) \backslash Z$. By the closeness of $Z$, for any $x \in X$ we can find $U_x$ open in $X$ and $V_x$ open in $Y$ such that $(x,y) \in U_x \times V_x$ and $(U_x \times V_x) \cap Z = \varnothing$. Now $X \times \{y\}$ is compact, so finitely many $U_{x_1} \times V_{x_1}, \dots, U_{x_n} \times V_{x_n}$ cover $X \times \{y\}$. Then $V = \bigcap_{i = 1}^n V_{x_n}$ is open in $Y$. Claim: $V \cap p(Z) = \varnothing$. Suppose not, say $y' \in V \cap p(Z)$. Then $(x', y') \in Z$ for some $x' \in X$. Since $U_{x_1}, \dots, U_{x_n}$ cover $X$, $x' \in U_{x_i}$ for some $i$. But then $(x', y') \in U_{x_i} \times V \subseteq U_{x_i} \times V_{x_i}$, a contradiction to $(U_{x_i} \times V_{x_i}) \cap Z = \varnothing$. Therefore we have found an open neighborhood $V$ of $y$ in $Y$ disjoint from $p(Z)$.

($\impliedby$) This direction is less trivial because of the construction of $Y$. Here I refer to Theorem 3.1.16 in Engelking's General Topology or Lemma 4.3 in Martin Escardó's note.

Let $\mathcal{U} = \{U_\alpha\}$ be an open cover of $X$ and $\widetilde{\mathcal{U}}$ be the set of all finite unions of sets in $\mathcal{U}$. To use the closed projection property, let $Y$ be the space whose points are open subsets of $X$, and define a subset $V \subseteq Y$ to be open if and only if (1) $V$ is upward-closed, i.e. if $U \in V$ and $U' \in Y$ satisfies $U \subseteq U'$ then $U' \in V$, and (2) if $X \in V$ then $V \cap \widetilde{\mathcal{U}} \neq \varnothing$. This defines a topology on $Y$, where the only nontrivial part is prove property (2) for the intersection $V_1 \cap V_2$ for $V_1, V_2$ satisfying (1) and (2): Suppose $X \in V_1 \cap V_2$ so we have $U_1 \in V_1 \cap \widetilde{\mathcal{U}}$ and $U_2 \in V_2 \cap \widetilde{\mathcal{U}}$. By the definition of $\widetilde{\mathcal{U}}$, we have $U_1 \cup U_2 \in \widetilde{\mathcal{U}}$. By property (1) of $U_1$ and $U_2$, we have $U_1 \cup U_2 \in V_1 \cap V_2$, so property (2) is proved. As a notation, for $U \in \widetilde{\mathcal{U}}$, let $V(U)$ be the set of all open subsets of $X$ containing $U$. Then $V(U)$ is clearly open in $Y$.

Let $Z = \{(x, U) \in X \times Y \,|\, x \not \in U\}$. Then $Z$ is a closed subset of $X \times Y$: Pick $(x, U) \in (X \times Y) \backslash Z$, so $x \in U$. Also we have $x \in U'$ for some $U' \in \widetilde{\mathcal{U}}$. Then $(U \cap U') \times V(U \cap U')$ is an open neighborhood of $X \times Y$ containing $(x, U)$, so $Z$ is closed. By the closed projection property, $p(Z)$ is closed in $Y$, so $Y \backslash p(Z) = \{U \in Y \,|\, (X \times \{U\}) \cap Z = \varnothing\}$ is open in $Y$. We see that $U \in Y \backslash p(Z)$ if and only if for any $x \in X$, $x \in U$, i.e. $X \subseteq U$. Then $X \in Y \backslash p(Z)$, so by property (2) we know that $(Y \backslash p(Z)) \cap \widetilde{\mathcal{U}} \neq \varnothing$. Then we have some $X \subseteq U$ for some finite union $U$ of sets in $\mathcal{U}$. Therefore $X$ is a compact topological space. ∎

We need to replace compactness by completeness for varieties because any variety is a noetherian topological space, hence compact.

Resuming to varieties, I want to compare the definition of complete varieties given by Hartshorne's Algebraic Geometry and the definition above. Hartshorne gave the following definitions in Sections II.3-II.4:

(i) An (abstract) variety $X$ is an integral separated scheme of finite type over an algebraically closed field $k$, i.e. $X$ is an integral scheme and we have a separated morphism $X \to \operatorname{Spec} k$ of schemes of finite type.
(ii) A morphism $f: X \to Y$ of schemes is closed iff it is closed as a map of topological spaces.
(iii) A morphism $f: X \to Y$ of schemes is universally closed iff for any morphism $Y' \to Y$ of schemes, the induced projection morphism $f': X \times_Y Y' \to Y'$ is also closed. (Here $X \times_Y Y'$ is the fibred product of $X$ and $Y'$ over $Y$, and the process $X \mapsto X \times_Y Y'$ is called a base extension by $Y' \to Y$.)
(iv) A morphism $f: X \to Y$ of schemes is proper iff it is separated, of finite type, and universally closed.
(v) An (abstract) variety $X$ is complete iff the morphism $X \to \operatorname{Spec} k$ is proper.

Comparing definitions (i), (iv) and (v), we see a lot of redundancy. In fact, the requirement of $X$ being complete says only that the morphism $X \to \operatorname{Spec} k$ is universally closed, that is, for any schemes $Y$ over $k$, the induced morphism $X \times_{\operatorname{Spec} k} Y \to Y$ is also closed. This agree with the definition in the Wikipedia page, if we are talking about abstract varieties.

For a (classical) quasi-projective variety $V$ over an algebraically closed field $k$, as defined in Chapter 1 of Hartshorne's book, we have the associated abstract variety $t(V)$ given by Propositions II.2.6 and II.4.10 We can recover the topological space $V$ by taking the set of closed points of $\operatorname{sp}(t(V))$. Similarly, we can relate the products of quasi-projective varieties over $k$ to fibred products of abstract varieties over $k$.

For convenience, we should denote the fibred product symbol $\times_{\operatorname{Spec} k}$ by $\times_k$.

Proposition. Let $X$ and $Y$ be schemes locally of finite type over $k$. Then there is an 1-1 correspondence between the set of pairs of closed points in the Cartesian product $X \times Y$ and the set of closed points of the fibred product $X \times_k Y$. Furthermore, for quasi-projective varieties $V$ and $W$, there is a homeomorphism between the product $V \times W$ and the subspace of closed points of the fibred product $t(V) \times_k t(W)$.

Proof. Recall that $\operatorname{Spec} k$ is a 1-point space whose structure sheaf is $k$. Recall also that if $X$ is locally of finite type over $k$, $x \in X$ is a closed point and $U$ is an open neighborhood of $x$ such that $(U, \mathscr{O}_X|_U)$ is an affine scheme, $(\varphi, \varphi^\#): (U, \mathscr{O}_X|_U) \to \operatorname{Spec} A$ being the isomorphism of locally ringed space with $A \simeq \mathscr{O}_X(U)$ a finitely generated $k$-algebra, then $x$ corresponds to a maximal ideal of $A$. Such a maximal ideal corresponds to a ring homomorphism $\phi_1: A \to k$ (because the quotient ring is a finite extension of $k$, so it has to be $k$).

Now let $y \in Y$ be a closed point of $Y$ and $(V, \mathscr{O}_Y|_V) \simeq \operatorname{Spec} B$ be an open affine neighborhood of $y$, with $B$ a finitely generated $k$-algebra, and $\phi_2: B \to k$ be the corresponding ring homomorphism. From the construction of the fibred product $X \times_k Y$ we see that $U \times_k V \simeq \operatorname{Spec} (A \otimes_k B)$ is an open affine subscheme of $X \times_k Y$. Inside $U \times_k V$ we have a closed point corresponding to the ring homomorphism $\phi_1 \otimes \phi_2: A \otimes_k B \to k$. Therefore from a pair of closed points $(x, y) \in X \times Y$, we have associated a closed point in $U \times_k V$.

Conversely, if we have a closed point in $U \times_k V$ with the corresponding ring homomorphism $\phi: A \otimes_k B \to k$, then composed with the natural map we get a pair of ring homomorphisms $\phi_1: A \to A \otimes_k B \to k$ and $\phi_2: A \to A \otimes_k B \to k$, which again corresponds to a pair of closed points $(x, y) \in X \times Y$. Clearly these two procedures are inverse to each other, hence we have a 1-1 correspondence, proving the first assertion. We can often write the corresponding closed point in $U \times_k V$ as $(x,y)$.

For quasi-projective varieties $V$ and $W$, the underlying set of the product $V \times W$ is the same as the Cartesian product, but the topology is defined using the Serge embedding, as in Exercise I.3.16. On the other hand, Proposition II.2.6 showed that the product $V \times W$ is homeomorphic to the set of closed points of $t(V \times W)$. Since $V \times W$ is the product in the category of varieties, $t(V) \times_k t(W)$ is the product in the category of schemes over $k$, and the fully faithful functor $t$ preserves products, we see that $t(V) \times_k t(W) \simeq t(V \times W)$ naturally as schemes over $k$ (This is Exercise II.3.23). Then we have the homeomorphism from the product $V \times W$ to the set of closed points of $t(V \times W)$, then to the set of closed points $t(V) \times_k t(W)$. ∎

Proposition. Let $V$ and $W$ be quasi-projective varieties over $k$. Then a morphism $f: V \to W$ of varieties is closed if and only if the associated morphism $t(f): t(V) \to t(W)$ is closed.

Proof. ($\impliedby$) This is trivial if we consider $V$ as a topological subspace of $t(V)$.

($\implies$) On the set level, the points of $t(V)$ are irreducible closed subsets of $V$. A continuous maps certainly sends irreducible subsets to irreducible subsets, and the closure of irreducible subsets are also irreducible, so a continuous map $f: V \to W$ determines the map $t(f): t(V) \to t(W)$ by $t(f)(Y) = \overline{f(Y)}$ if $Y$ is an irreducible closed subset of $V$.

The closed subsets of $t(V)$ are subsets of the form $t(Y)$, where $Y$ is a closed subset of $V$. If $t(Y)$ is a closed subset in $t(V)$, then $Y$ is closed in $V$, so by assumption $Z = f(Y)$ is closed in $W$. We need to show that $t(Z)$ is the image of $t(Y)$ under $t(f)$. If $Y'$ is an irreducible closed subset of $Y$, then $f(Y') \subseteq Z$ so $t(f)(Y') = f(Y') \in t(Z)$. This shows that $t(f)$ maps $t(Y)$ into $t(Z)$. To show that it is surjective, I will use noetherian induction on the noetherian topological space $W$. (See Exercise II.3.16.) Let $\mathscr{P}$ be the property of closed subsets $Z$ that $t(f): t(Y \cap f^{-1}(Z)) \to t(Z)$ is surjective. Then it is vacuously true when $Z = \varnothing$. Now assume $\mathscr{P}$ holds for all proper closed subsets $Z'$ of $Z$. Let $Z'$ be an irreducible closed subset of $Z$. If $Z' = Z$, then $Y \in t(Y)$ and $t(f)(Y) = f(Y) = Z$. If $Z' \neq Z$, then by the inductive hypothesis, $t(f): t(Y \cap f^{-1}(Z')) \to t(Z')$ is surjective. Since $Z' \in t(Z')$, we have some irreducible closed subset $Y'$ of $Y \cap f^{-1}(Z')$ such that $t(f)(Y') = f(Y') = Z'$, and $Y' \in t(Y)$. This shows that $t(f): t(Y) \to t(Z)$ is surjective. Hence $t(f)$ maps $t(Y)$ to the closed subset $t(Z)$. ∎

Using $t(V) \times_k t(W) \simeq t(V \times W)$ again, we obtain the following corollary.

Corollary. Let $V$ and $W$ be quasi-projective varieties over $k$. Then $V \times W \to W$ is closed if and only if the associated morphism $t(V) \times_k t(W) \to t(W)$ of schemes over $k$ is closed. ∎

Therefore, one sees that a quasi-projective variety $V$ is complete if and only if the associated abstract variety $t(V)$ is complete.

Proposition. If $X$ is a complete abstract variety and $Z$ is a closed subscheme of $X$, then $Z$ is proper over $k$. In particular, if $Z$ is an irreducible closed subset of $X$ and $Z$ has the reduced induced closed subscheme structure, then $Z$ is a complete abstract subvariety of $X$.

Proof. We have a closed immersion $f: Z \to X$ where $Z$ has the closed subscheme structure. By Collary II.4.8, $f$ is proper and $X \to \operatorname{Spec} k$ is proper by definition, so the composition $Z \to X \to \operatorname{Spec} k$ is proper. In case $Z$ is both reduced and irreducible, then $Z$ is also integral, so $Z$ is an abstract variety. ∎

Proposition. If $X$ is a complete abstract variety, then any morphism $f: X \to Y$ from $X$ to an abstract variety is closed.

Proof. Since $X$ is a noetherian topological space, any closed subset of $X$ is a finite union of irreducible closed subsets of $X$. Therefore it suffices to show that the image of an irreducible closed subset of $X$ is closed in $Y$.

Let $Z$ be an irreducible closed subset of $X$. Give $Z$ the reduced induced closed subscheme structure, so $Z$ is now a complete abstract variety. The identity morphism $Z \to Z$ and $f: Z \to Y$ induces the graph morphism $\Gamma_f: Z \to Z \times_k Y$ with $p_1 \circ \Gamma_f = \operatorname{id}_Z$ and $p_2 \circ \Gamma_f = f$, where $p_1, p_2$ are projections of the fibred product $Z \times_k Y$. The morphisms $f \circ p_1: Z \times Y \to Z$ and $p_2: Z \times_k Y \to Y$ induce $g: Z \times_k Y \to Y \times_k Y$ with $\pi_1 \circ g = f \circ p_1$ and $\pi_2 \circ g = p_2$, where $\pi_1, \pi_2$ are projections of the fibred product $Y \times_k Y$. Intuitively, we think of $\Gamma_f$ as $(\operatorname{id}_Z, f)$ and $g$ as $(f, \operatorname{id}_Y)$. We see that $f: Z \to Y$ can be factorized as

$Z \overset{\Gamma_f}{\longrightarrow} Z \times_k Y \overset{g}{\longrightarrow} Y \times_k Y \overset{\pi_i}{\longrightarrow} Y$

where $\pi_i$ can be $\pi_1$ or $\pi_2$. Since $Y$ is separated over $k$, the diagonal morphism $\Delta: Y \to Y \times_k Y$ is a closed immersion, and its image $\Delta(Y)$ is closed in $Y \times_k Y$, so $g^{-1}(\Delta(Y))$ is closed in $Z \times_k Y$. We see that on the set level the image $f(Z) = p_2(\Gamma_f(Z)) = p_2(g^{-1}(\Delta(Y))$ is the image of a closed subset of $Z \times_k Y$ under the projection $p_2: Z \times_k Y \to Y$, so it is closed in $Y$ by the completeness of $Z$. ∎

The next proposition says that a quasi-projective variety $V$ is complete if and only if it is projective.

Proposition. Let $X$ be a quasi-projective abstract variety over an algebraically closed field $k$. Then $X$ is complete if and only if $X$ is projective.

Proof. ($\impliedby$) Since $X$ is of finite type over $k$, $X$ can be covered by finitely many open affine subchemes $(U_i, \mathscr{O}_X|_{U_i}) \simeq \operatorname{Spec} A_i$ where each $A_i$ is a finitely generated $k$-algebra, so $X$ is a noetherian scheme. Then this direction follows from Theorem II.4.9, applied to the morphism $X \to \operatorname{Spec} k$.

($\implies$) By the definition of being quasi-projective, the morphism $X \to \operatorname{Spec} k$ can be factorized into an open immersion $j: X \to X'$ to some scheme $X'$ and a projective morphism $i: X' \to \operatorname{Spec} k$. Then $X'$ is a noetherian scheme. Since $j(X)$ is irreducible, by replacing $X'$ by a closed subscheme $Y'$ whose underlying topological space is an irreducible component of $\operatorname{sp}(X')$, we can assume that $X'$ is irreducible. Since $X$ is complete, the morphism $j: X \to X'$ is closed by the last proposition. But if $j(X)$ is a proper open subset of $X'$, then $j(X')$ cannot be closed in $X'$ because $X'$ is irreducible, a contradiction. So $j(X) = X'$ and $j$ is an isomorphism between $X$ and $X'$, so $X$ is projective. ∎

Finally, let me mention several facts that may be my next post on the topic.

(1) The class of abstract varieties is larger than the class of quasi-projective abstract varieties, and there exists a non-projective complete abstract variety.
(2) On the other hand, Chow's Lemma states that for any complete abstract variety $X$, there exists a projective abstract variety $X'$ and a surjective birational morphism $X' \to X$.

Pizza Seminar: Which integral is elementary?

Here is a self-reminding note of my graduate student seminar (a.k.a. Pizza seminar) at Rutgers.

Title: Which integral is elementary?

Abstract: It is "well known" that the indefinite integrals of $e^{-x^2}$ and $1/\log x$ are not elementary functions, but the reason behind is usually a blindspot for many calculus or undergraduate classes. In this talk, I will discuss two theorems due to Liouville that characterize the types of integrands whose integrals are elementary functions.

Link to PDF

Some Hartshorne's Exercises

Link: https://drive.google.com/file/d/0B--hvWgzOhnJbUFUOTdya25KdEU/view?usp=sharing

The label ****** means incomplete.

The Long Line

As we learnt in undergraduate courses, the real line is a linear continuum, i.e. it is a totally (linearly) ordered set $X$ that satisfies the axioms:

(i) For any $x, y \in X$, if $x < y$, then there exists $z \in X$ such that $x < z < y$.
(ii) (Supremum property) Any nonempty subset $A$ of $X$ that is bounded above has a least upper bound, i.e. an element $s \in X$ such that $a < s$ for all $a \in A$ and if $x < s$ then $x$ is not an upper bound of $A$. (A least upper bound is also called a supremum.)

It can be shown that a linear continuum with more than one point has at least uncountably many points.

A nonempty perfect set in $\mathbb{R}$ which contains no rational number

When I am doing some measure theory exercises today, I suddenly think of this problem in chapter 2 of Rudin's Principles of Mathematical Analysis which I found difficult in my first year undergraduate study but seems easy to me now. I want to present the construction here.

We want to imitate the construction of the Cantor set so that we obtain a perfect set. Fix some closed interval $[a,b]$ with irrational endpoints and let $\{q_0, \;q_1, \;q_2, \dots\}$ be an enumeration of the rational points in $[a,b]$. We will remove all the $q_n$ from the interval $[a,b]$ in the process of constructing a decreasing sequence of subsets $S_n$ satisfying the properties: (i) $S_n$ is compact, (ii) $S_n$ has nonempty interior, (iii) the boundary of $S_n$, denoted by $\partial S_n$, consists of irrational numbers and (iv) $\partial S_n \subset \partial S_{n+1}$.

We do this: Set $S_0=[a,b]$. Then $S_0$ satisfies properties (i), (ii) and (iii). To construct $S_{n+1}$ from a given $S_n$ which satisfies assumption (i), (ii) and (iii), we find the least $k$ such that $q_k\in S_n$ (which exists by assumption (ii)). By assumption (iii), $q_k$ is not on the boundary, hence is in the interior of $S_n$. So we can find an open interval $(r,s)$ with irrational endpoints lying in the interior of $S_n$, i.e. $r, s \in {S_n}^{\circ} \backslash \mathbb{Q}$. Set $S_{n+1} = S_n \backslash (r,s)$. We see that $S_{n+1}$ is compact by assumption (i), $S_n$ has nonempty interior because some small interval $(r-\delta, r] \subseteq S_{n+1}$ and $\partial S_{n+1} = \partial S_n \cup \{r,s\}$ consists of irrational points. So assumptions (iii) and (iv) hold for $S_{n+1}$.

Let $S = \cap_{n=0}^\infty S_n$. We claim that $S$ is the desired set: $S$ is nonempty and closed as an intersection of nonempty compact sets. Since at least one rational number is removed from $S_n$ in each step, $q_n \notin S_{n+1}$ hence $S$ contains no rational number. Also, every point in $S$ is a limit point of $S$. To see this, notice that $\partial S_n \subseteq S$ for every $n$. If $x \in S$ is an irrational point which is not a right endpoint of any $S_n$, then for any given $\epsilon > 0$ we can find a rational $q \in [a,b]$ such that $x < q$ and $|x-q| < \epsilon$. Consider the interval $(r,s)$ which contains $q$ being removed in some step. We have $x \leqslant r < q$, and $x \neq r$ because $x$ is not a right endpoint, then $r \in \partial S_n \subseteq S$ for some $n$ and $|x-r| < \epsilon$. Note that by our construction a point cannot be a right endpoint and a left endpoint at the same time, so if $x$ happens to be a right endpoint we can take $q < x$ and consider $q < s < x$, so we have proved that every $x \in S $ is a limit point of $S$. Therefore $S$ is the desired perfect set.

Dimension of infinite dimensional vector spaces

A vector space is said to be infinite dimensional iff it cannot be spanned by a finite set of vectors. Unlike the finite dimensional cases, not much is said about the dimension of infinite dimensional spaces in typical linear algebra courses. Therefore we supplement the analogous facts here, including that the dimension of an infinite dimensional space is well-defined. The following are some examples of infinite dimensional vector space:

Example.
(a) The set of real polynomials in one variable $x$ \[ \mathbb{R}[x] = \{a_n x^n + \dots + a_1 x + a_0 | n \in \mathbb{N} \text{ and } a_i \in \mathbb{R} \text{ for all } i \in \{0, ..., n\}\} \] as a real vector space has a countable basis, the set of monomials together with the constant polynomial 1, $\{1, x, x^2, x^3, \dots\}$.

(b) The set of infinite $\mathbb{F}_2$ sequences \[\mathbb{F}_2^\mathbb{N}\ = \{(b_0, b_1, b_2, \dots)|b_i \in \mathbb{F}_2 \text{ for all }i \in \mathbb{N}\}\]as an $\mathbb{F}_2$ vector space cannot be spanned by a countable subset. To see this, notice that the vector space is uncountable but the span of a countable subset using a finite field is countable. \[\operatorname{span}_{\mathbb{F}_2}(\{v_1, v_2, \dots\}) = \cup_{k=1}^{\infty}{\{\sum_{i=1}^{k}{c_i v_i}}|c_i \in \mathbb{F}_2 \text{ for all } i \in \{1,2,\dots,k\}\}.\]

(c) The set of infinite real number sequences \[\mathbb{R}^\mathbb{N}\ = \{(a_0, a_1, a_2, \dots)|a_i \in \mathbb{R} \text{ for all }i \in \mathbb{N}\}\] and the set of bounded functions on the real line \[B(\mathbb{R}) = \{f:\mathbb{R} \rightarrow \mathbb{R} | \operatorname{sup}_{x \in \mathbb{R}}{|f(x)|} < \infty\}\] as real vector spaces cannot be spanned by a countable subset. This is a consequence of the Baire Category Theorem in functional analysis. ∎

To work with infinite dimensional spaces, it is more convenient to characterize bases in a different way. A linearly independent subset is called maximal iff it is not a proper subset of any linearly independent subset.

Proposition. A subset is a basis iff it is a maximal linearly independent subset.

Proof. Let $S$ be a linearly independent subset of a vector space $V$. $S$ does not span $V$ iff there exists $v\in V-\operatorname{span}(S)$ iff there exists $v\in V$ such that $S\cup \{v\}$ is linearly independent iff $S$ is not maximal. ∎

We can find a maximal linearly independent subset in any vector space using the Zorn's lemma, which is logically equivalent to the axiom of choice in set theory.

Zorn's Lemma. If $\mathcal{S}$ is a nonempty partially ordered set in which every chain (totally ordered subset) has an upper bound in $\mathcal{S}$, then $\mathcal{S}$ has at least one maximal element. ∎

Theorem. Every vector space has a basis.

Proof. Let $\mathcal{S}$ be the set of linearly independent subsets in the vector space, partially ordered by inclusion. ($\varnothing \in \mathcal{S}$ so $\mathcal{S}$ is nonempty.) We will show, by Zorn's Lemma, that $\mathcal{S}$ has a maximal element, i.e. a maximal linearly independent subset. Then this subset is a basis for the vector space.

We verify the hypothesis of Zorn's Lemma. Given a chain $C = \{S_\alpha|\alpha \in I\}$ in $\mathcal{S}$, we show that $S_I = \cup_{\alpha\in I}{S_\alpha}$ is linearly independent, hence an upper bound of $C$. For any finite subset $\{v_1, v_2, \dots, v_n\}$ of $S_I$, there are $\alpha_1, \alpha_2, \dots, \alpha_n \in I$ such that $v_i \in S_{\alpha_i}$. Since $\{S_{\alpha_i}\}$ is totally ordered, there is a greatest element $S_\alpha$ among them, i.e. $S_{\alpha_i}\subseteq S_\alpha$ for all $i$. Then $\{v_1, v_2, \dots, v_n\}\subseteq S_\alpha$ is a linearly independent subset. By definition, $S_I$ is linearly independent. ∎

The next proposition is a slight modification of the theorem.

Proposition.
(a) Every linearly independent subset can be extended to a basis.
(b) Every spanning subset contains a basis.

Proof. Let $S$ be a spanning subset and $L$ be a linearly independent subset.
(a) In the previous proof, replace $\mathcal{S}$ by the set of linearly independent subsets containing $L$.
(b) Replace $\mathcal{S}$ by the set of linearly independent subsets contained in $S$ and obtain a maximal element $M$ (which is maximal in the sense that it is not a proper subset of another linearly independent subset contained in $S$). Assume that $M$ is not a maximal linearly independent subset, then there exists $v \in V-\operatorname{span}(M)$. Write $v=c_1 v_1 + c_2 v_2 + \dots + c_n v_n$ where $\{v_1, v_2, \dots, v_n\}\subseteq S$. Some $v_i \not\in \operatorname{span}(M)$, say $v_1$, then $M \cup \{v_1\}$ is a linearly independent subset strictly containing $M$, contradicts the maximality of $M$. ∎

A result in cardinal arithmetic is needed to proceed.

(Cardinals (or cardinal numbers) are a measure of the cardinality of sets. A cardinal is a natural choice of set that represent all sets with the same cardinality as it. To be more specific, $\operatorname{card}(A)$ is a cardinal that has the same cardinality as the set $A$.)

Proposition. If $\kappa$ is an infinite cardinal, then $\kappa\cdot\kappa=\kappa$, where $\kappa\cdot\kappa$ is defined to be $\operatorname{card}(\kappa\times\kappa)$. ∎

In other words, if a set $A$ is infinite, then $\operatorname{card}(A \times A) = \operatorname{card}(A)$. The fact that $\mathbb{N} \times \mathbb{N}$ is countable is included as a special case.

Remark. Writing the Cartesian product as a union, it follows that the union of at most $\kappa$ many sets, each of cardinality at most $\kappa$, has cardinality at most $\kappa$. Also, apply the proposition several times we get $\kappa^n = \kappa$ for all $n \in \mathbb{N}-\{0\}$.∎

Here comes the main fact:

Theorem. The cardinality of a spanning subset has at least the cardinality of a linearly independent subset.

Proof. Let $S$ be a spanning subset and $L$ be a linearly independent subset of a vector space $V$. What we need to show is $\operatorname{card}(S)\geq\operatorname{card}(L)$.

Extend $L$ to a basis $L'$, then $\operatorname{card}(L) \leq \operatorname{card}(L')$. Every $v \in S$ can be expressed as a linear combination of some finite subset $L'_v$ of $L'$, so $V = \operatorname{span}(S) \subseteq \operatorname{span}(\cup_{v \in S}{L'_v})$. As $L'$ is a maximal linearly independent subset, no proper subset of it can span $V$, hence $\cup_{v \in S}{L'_v} = L'$. By the remark, $\operatorname{card}(L) \leq \operatorname{card}(L') = \operatorname{card}(\cup_{v \in S}{L'_v}) \leq \operatorname{card}(S)$. ∎


Corollary. Any two bases for a vector space has the same cardinality.

Proof. Apply the previous theorem twice and the Cantor-Bernstein-Schröder Theorem in set theory. ∎

Finally, we include two more results. The first generalizes the technique used in example (b), which relates the cardinality and the dimension of a vector space. The second constructs vector spaces of arbitrary dimension and describes all vector spaces.

Proposition. Let $V$ be a vector space over a field $F$ and $\operatorname{dim}V$ be its dimension over $F$.
(a) If $V$ is finite dimensional, then $\operatorname{card}(V) = \operatorname{card}(F)^{\operatorname{dim}V}$. In particular, if $F$ is infinite, then $\operatorname{card}(V) = \operatorname{card}(F)$.
(b) If $V$ is infinite dimensional, then $\operatorname{card}(V) = \max{\{\operatorname{card}(F), \operatorname{dim}V\}}$.

Proof.
(a) The first part is just typical linear algebra, and the second part follows from $\operatorname{card}(F^{\operatorname{dim}V}) = \operatorname{card}(F)^{\operatorname{dim}V} = \operatorname{card}(F)$ by the remark.
(b) Let $B$ be a basis for $V$. Then $V$ is in one-one correspondence with the set of finite subsets of $(F - \{0\}) \times B$. By the remark, since $(F-\{0\}) \times B$ is infinite (as $B$ is), the latter set has cardinality $\operatorname{card}((F-\{0\}) \times B)$, which equals $\max{\{\operatorname{card}(F-\{0\}), \operatorname{card}(B)\}} = \max{\{\operatorname{card}(F), \operatorname{card}(B)\}}$$= \max{\{\operatorname{card}(F), \operatorname{dim}V\}}$ by the previous proposition. ∎

Definition. Given a field $F$ and an arbitrary set $X$. The set of functions from $X$ to $F$ which vanishes outside a finite subset of $X$, \[(F^X)_0 = \{f: X \rightarrow F | f^{-1}(F-\{0\}) \text{ is finite}\},\]is a vector space over $F$ and is called a coordinate space. ∎

The polynomial ring in one variable is isomorphic to $(F^\mathbb{N})_0$ as vector space.

Proposition. Given a field $F$.
(a) The coordinate space $(F^X)_0$ has dimension $\operatorname{card}(X)$ over $F$. Hence there exist vector spaces over $F$ of arbitrary dimension.
(b) Every vector space over $F$ is isomorphic to a coordinate space over $F$.

Proof.
(a) $(F^X)_0$ has a canonical basis $\{f_x\}_{x \in X}$ where $f_x(x) = 1$ and $f_x(y) = 0$ for all $y \in X -\{x\}$. Given an arbitrary cardinal $\kappa$, the vector space $(F^\kappa)_0$ has dimension $\kappa$.
(b) Let $V$ be a vector space over $F$ and $B$ be a basis for $V$. The linear map $\Phi: V \rightarrow (F^B)_0$ which sends $v \in B$ to $f_v \in (F^B)_0$ is an isomorphism. ∎

Welcome to my blog!

The site aims to collect well-known, remarkable but scattered mathematical facts and theorems of interest to pre-university or non-mathematics major audience. An introduction to modern mathematics in logical order is also under planning. Thanks Alexan Fung, one of my university schoolmates, for giving the name to this blog hence freeing me from this bothersome matter. The mathematical symbols are written using MathJax (Getting Started).