Tangents (heh) on Cardinality

Lee Smooth Manifolds

I thought, well, since I have way more time in my hands than I know what to do with, I might as well start to write down full solutions for all (or at least all the nontrivial) exercises in Chapter 11 of Lee. It ended up sorta devolving into a huge tangent into cardinalities because for some reason 11.2 was very technically difficult.

Problem 1 (11.1). (a) Suppose $V$ and $W$ are finite-dimensional vector spaces and $A:V\to W$ is any linear map. Show that the following diagram commutes: where ${\xi }_V$ and ${\xi }_W$ denote the canonical isomorphism with the double-dual space. (b) Show that there does not exist a way to assign to each finite-dimensional vector space $V$ an isomorphism $\beta :V\to V^{\ast }$ such that for every linear map $A:V\to W$, the following diagram commutes:

Proof) (a) is really just an annoying spell-out-definitions exercise. For any $p\in V$ consider $q=(A^{\ast }{)}^{\ast }\circ {\xi }_V(p)$. For any $y\in W^{\ast }$, we have the following:

$$qy=((A^{\ast }{)}^{\ast }\circ {\xi }_V(p))(y)={\xi }_V(p)(A^{\ast }(y))=A^{\ast }(y)(p)=y(Ap)={\xi }_W(Ap)y\Rightarrow q={\xi }_W(Ap)$$

(b) Setting $A=0$ gives an immediate contradiction.

Before proceeding to 11.2 we prove some fundamental results on cardinalities; we keep to set-theory basics for now and don’t reach into more complex cardinality arithmetic.

Theorem 2. If $X$ is infinite and $F=\{f_1,\dots ,f_n\}$ is finite, $|X|=|X\bigsqcup F|$ where $\bigsqcup$ denotes disjoint union.

Proof) Clearly, $|X\bigsqcup F|\ge |X|$; it suffices to show $|X|\ge |X\bigsqcup F|$. Let $d:\mathbb{N}\to X$ be an injection. Define a map $\varphi :X\bigsqcup F\to X$ as follows: if $x\in d(\mathbb{N})$, $\varphi (x)=d(d^{-1}(x)+n)$, if $x\in X\setminus d(\mathbb{N})$, $\varphi (x)=x$, and for all $f_i\in F$, $\varphi (f_i)=d(i)$. This is clearly injective, and we are done.

Theorem 3. If $X$ is infinite, $|X|\cdot {\aleph }_0=|X\times \mathbb{N}|=|X|$.

Proof) The theorem trivially holds if $|X|={\aleph }_0$; thus, a natural motivation is to partition $X$ into countably infinite sets and apply piecewise. Consider all partitions of subsets of $X$ into countably infinite subsets, with a partial order given by $\mathcal{A}\le \mathcal{B}$ iff $\mathcal{A}\subset \mathcal{B}$ (as families). We can apply Zorn’s lemma here to obtain a maximal element $\mathcal{C}$; if $X\setminus {\bigcup }_{C\in \mathcal{C}}C$ is infinite we can find a countably infinite subset and adjoin it to $\mathcal{C}$ to obtain a larger family, and thus $X\setminus {\bigcup }_{C\in \mathcal{C}}C$ is finite. By replacing an arbitrary $C\in \mathcal{C}$ with $C\cup (X\setminus {\bigcup }_{C\in \mathcal{C}}C)$, we obtain our desired partition. For any $C\in \mathcal{C}$, let ${\varphi }_C:C\times \mathbb{N}\to C$ be a bijection, and let $\varphi :X\times \mathbb{N}\to C$ be the union of the ${\varphi }_C$s. This is clearly bijective, and thus we are done.

Remark: A free corollary of Theorem 3 is that, for any finite set $F$, $|X|\cdot |F|=|X\times F|=|X|$.

Theorem 4. $\kappa \times \kappa =\kappa$ for every infinite cardinal $\kappa$.

Proof) This is just a fancy way of saying $|A\times A|=|A|$ for every infinite $A$. We proceed by considering the collection

$$P=\{(S,f):S\subset A,f:S\times S\to S\text{ is bijective}\}$$

and impose a partial order $\ge$ on $P$ given by $(S,f)\ge (S^{\prime },f^{\prime })$ if $S^{\prime }\subset S$ and $f^{\prime }$ agrees with $f$ on $S\times S$. This set is nonempty as ${\aleph }_0\cdot {\aleph }_0={\aleph }_0$ and $A$ has a countably infinite subset. Clearly, any chain $\{(S_i,f_i){\}}_{i\in I}$ has an upper bound $({\bigcup }_{i\in I}{S}_i,f={\bigcup }_{i\in I}{f}_i)$ which is well-defined, and by Zorn’s lemma there is a maximal element $(S,f)$ of $P$. We claim that $|S|=|A|$, which finishes the proof. FTSOC assume otherwise.

First, note that clearly $|S|<|A\setminus S|$, as otherwise $|A|=|S\cup (A\setminus S)|\le |S\times \{1,2\}|=|S|$, forcing $|S|=|A|$. Therefore, there is some subset $T\subset A\setminus S$ such that $|T|=|S|$. We can then extend $(S,f)$ to $S\cup T$ as follows: defining $D=(S\times T)\cup (T\times S)\cup (T\times T)$, we can see that $|D|\le |(S\times S)\times \{1,2,3\}|=|S\times S|=|S|=|T|$, and thus there is a bijection $h:D\to T$. Define $\tilde{f}:(S\cup T)\times (S\cup T)\to S\cup T$ as $\tilde{f}=f$ on $S\times S$ and $\tilde{f}=h$ on $D$. Then, $\tilde{f}$ is our desired bijection, contradicting the maximality of $f$.

Theorem 5 (EB.5). Suppose $V$ is an infinite-dimensional vector space. Any two bases for $V$ have the same cardinality.

Proof) Let $S$ and $T$ be two bases for $V$, and, FTSOC, assume $|S|>|T|$. Any $t\in T$ can be uniquely represented as a finite linear combination $t=\sum s_i$; let $S_t\subset S$ be the finite subset of elements appearing in the representation of $t$. Take $S_0={\bigcup }_{t\in T}{S}_t$; it suffices to show that $|S_0|<|S|$, as any element $s\in S\setminus S_0$ can be represented as a finite linear combination of elements of $S_0$, contradicting linear independence.

First note there exists an injection $\varphi :S_0\to T\times \mathbb{N}$ as follows; well-order $T$ and $S_0$, and let $s\in S_0$ map to $(t,k)$ where $t\in T$ is the smallest $t$ for which $s\in S_t$, and $s$ is the $k$th smallest element in $S_t$. Therefore, $|S_0|\le |T|\cdot {\aleph }_0=|T|<|S|$ by Theorem 3, and we are done.

Problem 6 (11.2). Suppose $V$ is an infinite-dimensional vector space and $V^{\ast }$ is its algebraic dual space. (a) Prove that there is an injective linear map from $V$ to $V^{\ast }$. (b) Prove that there is no injective linear map from $V^{\ast }$ to $V$. (c) Show there cannot be an isomorphism between $V$ and $V^{\ast \ast }$.

Proof) Let $A=\{v_i\}\subset V$ be a basis for $V$. Define a map $f:V\to V^{\ast }$ such that $f(v_i)$ is the element that sends $v_i$ to $1$ and all other $v_j$s to $0$; this is clearly an injective linear map from $V$ to $V^{\ast }$. (b) requires a much more intricate argument; it suffices to show there is a linearly independent set of $V^{\ast }$ that has cardinality equal to the power set of $A$ (all linearly independent sets can be extended to a basis by Zorn’s lemma, so this finishes the proof). We construct this as follows: first, let $\text{Fin}(A)$ be the collection of finite subsets of $A$; clearly, $|A|\le \text{Fin}(A)$. Furthermore, note that the set ${\text{Fin}}_n(A)$ of size-$n$ subsets of $A$ ($n$ is finite) satisfies $|{\text{Fin}}_n(A)|\le |A^n|=|A|$ by Theorem 4, and thus $|\text{Fin}(A)|=|{\bigcup }_{n\in \mathbb{N}}{\text{Fin}}_n(A)|\le |A\times \mathbb{N}|=|A|$. We have therefore shown that $|\text{Fin}(A)|=|A|$.

Let $\varphi :\text{Fin}(A)\to A$ be this bijection. Define $F:\mathcal{P}(A)\to V^{\ast }$ such that, for any $S\subset A$, $F(S)$ is the functional that satisfies $F(S)\varphi (S^{\prime })=1$ if $S^{\prime }\subset S$ and $0$ otherwise. Assume some nonzero-coefficient linear combination of $F(S_i)$ for $1\le i\le n$ is zero, FTSOC. Ordering by set inclusion, there must be some maximal element that we, WLOG, assume is $S_1$. For any $i\ge 2$ there must be some element $s_i\in S_1\setminus S_i$; taking the finite subset $U=\{s_i{\}}_{2\le i\le n}$ of $S_1$ we can see $F(S_1)$ is such that the basis element $\varphi (U)$ maps to $1$ while, for all $i\ge 2$, $F(S_i)$ maps $\varphi (U)$ to $0$. This is a clear contradiction, and thus all the $F(S_i)$s are linearly independent (this also cleanly implies injectivity). Thus, we are done as the image of $F$ is a linearly independent set of $V^{\ast }$ with cardinality equal to $\mathcal{P}(A)$, which is strictly greater than $A$.

The proof of (c) easily follows from (a) and (b) as the cardinality of the basis of $V^{\ast }$ is strictly greater than that of $V$; therefore, the cardinality of the basis of $V^{\ast \ast }$ is strictly greater than $V$, making an isomorphism between these spaces impossible.