04. 기저의 존재성Existence of a Basis

Solution. Classifying the subsets of $S = \{1,2,3,4\}$ by the parity of the sum of their elements, the members of $\mathcal{F}$ are exactly the following eight sets.

$$\{1\},\ \{3\},\ \{1,2\},\ \{1,4\},\ \{2,3\},\ \{3,4\},\ \{1,2,4\},\ \{2,3,4\}.$$

(The empty set and $S$ itself have element sums $0$ and $10$, respectively, both even, so neither belongs to $\mathcal{F}$.)

1. Among these eight sets, $\{1\}, \{1,2\}, \{1,4\}$ are each contained in $\{1,2,4\}$, and $\{3\}, \{2,3\}, \{3,4\}$ are each contained in $\{2,3,4\}$. The only member of $\mathcal{F}$ that could properly contain $\{1,2,4\}$ or $\{2,3,4\}$ is $S$ itself, and $S \notin \mathcal{F}$; hence $\{1,2,4\}$ and $\{2,3,4\}$ are each maximal elements of $\mathcal{F}$. Thus $\mathcal{F}$ has exactly two maximal elements, $\{1,2,4\}$ and $\{2,3,4\}$.
2. Every element of $\mathcal{F}$ has size $1$, $2$, or $3$, since the empty set and $S$ itself are excluded. In a chain, distinct sets are pairwise comparable, so their sizes are strictly increasing; as only three sizes are available, a chain in $\mathcal{F}$ has at most $3$ elements. The chain $$\{1\} \subset \{1,2\} \subset \{1,2,4\}$$ has $3$ elements in $\mathcal{F}$, so it is a chain with the largest possible number of elements.

Solution. Consider $\mathcal{F} = \{\{n\} : n \in \mathbb{N}\}$, the family of all singleton subsets of $\mathbb{N}$.

$\mathcal{F}$ is infinite, since it contains a distinct element $\{n\}$ for each natural number $n$. If $n \neq m$, then neither $\{n\} \subset \{m\}$ nor $\{m\} \subset \{n\}$ holds, so any two distinct elements of $\mathcal{F}$ are incomparable. Hence any chain in $\mathcal{F}$ can contain at most one element, and in particular every chain in $\mathcal{F}$ is finite.

Proof. Let $\mathcal{F} = \{A \subset S_2 : S_1 \cup A \text{ is linearly independent}\}$, partially ordered by inclusion.

Since $S_1$ is linearly independent, $S_1 \cup \emptyset = S_1$ is linearly independent, so $\emptyset \in \mathcal{F}$ and $\mathcal{F} \neq \emptyset$.

Let $\mathcal{C}$ be an arbitrary chain in $\mathcal{F}$. If $\mathcal{C} = \emptyset$, then $\emptyset \in \mathcal{F}$ is an upper bound for $\mathcal{C}$. Otherwise, let $U = \bigcup_{A \in \mathcal{C}} A$; we show $U \in \mathcal{F}$. Let $\{v_1, \ldots, v_n\}$ be any finite subset of $S_1 \cup U$ and suppose $\sum_{i=1}^n c_i v_i = 0$. For each $i$ with $v_i \in U$, choose $A_i \in \mathcal{C}$ with $v_i \in A_i$; only finitely many such $A_i$ occur, and since $\mathcal{C}$ is a nonempty chain, one of them—say $A_{i_0}$—contains all the others (if no $v_i$ lies in $U$, let $A_{i_0}$ be any member of $\mathcal{C}$). Then $\{v_1, \ldots, v_n\} \subset S_1 \cup A_{i_0}$: each $v_i \in U$ lies in some $A_i \subset A_{i_0}$, and each $v_i \in S_1$ lies in $S_1$. Since $A_{i_0} \in \mathcal{C} \subset \mathcal{F}$, the set $S_1 \cup A_{i_0}$ is linearly independent. Hence $c_1 = \cdots = c_n = 0$. As every finite subset of $S_1 \cup U$ is independent, $S_1 \cup U$ is linearly independent, so $U \in \mathcal{F}$, and $U$ is an upper bound for $\mathcal{C}$.

Thus every chain in $\mathcal{F}$ has an upper bound in $\mathcal{F}$, so by Zorn's Lemma $\mathcal{F}$ has a maximal element $S_2'$. Let $\beta = S_1 \cup S_2'$; since $S_2' \in \mathcal{F}$, $\beta$ is linearly independent.

We show $\beta$ generates $V$. Since $S_2$ generates $V$, it suffices to show $S_2 \subset \operatorname{span}(\beta)$. Suppose not; then there is $v \in S_2$ with $v \notin \operatorname{span}(\beta)$. Note $v \notin S_2'$, since $S_2' \subset \beta \subset \operatorname{span}(\beta)$. If $\beta \cup \{v\}$ were linearly dependent, then, since $\beta$ is independent, the coefficient of $v$ in the dependence relation would have to be nonzero, which would give $v \in \operatorname{span}(\beta)$, a contradiction. Hence $\beta \cup \{v\} = S_1 \cup (S_2' \cup \{v\})$ is linearly independent, and since $S_2' \cup \{v\} \subset S_2$, this means $S_2' \cup \{v\} \in \mathcal{F}$. But $S_2' \subsetneq S_2' \cup \{v\}$, contradicting the maximality of $S_2'$ in $\mathcal{F}$.

Therefore $S_2 \subset \operatorname{span}(\beta)$, and $V = \operatorname{span}(S_2) \subset \operatorname{span}(\beta) \subset V$, so $\beta$ generates $V$. Since $\beta = S_1 \cup S_2'$ is linearly independent and generates $V$, it is a basis for $V$.

Proof.

Note. The second "$W$" in the statement appears to be a typo for "$V$"; the proof below extends a basis of $W$ to a basis of $V$.

1. Let $\beta_W$ be a basis for $W$. Then $\beta_W \subset W \subset V$ and $\beta_W$ is linearly independent. Since $V$ itself trivially generates $V$, applying EX 2 with $S_1 = \beta_W$ and $S_2 = V$ gives a subset $S_2'$ of $V$ such that $\beta_W \cup S_2'$ is a basis for $V$. This extends $\beta_W$ to a basis for $V$.
2. Since $S_1$ is linearly independent and $S_2$ generates $V$, EX 2 gives a subset $S_2'$ of $S_2$ such that $\beta = S_1 \cup S_2'$ is a basis for $V$. As $S_1 \subset S_2$ and $S_2' \subset S_2$, we have $\beta = S_1 \cup S_2' \subset S_2$, and also $S_1 \subset \beta$. Hence there is a basis $\beta$ of $V$ with $S_1 \subset \beta \subset S_2$.
3. Since $\beta$ is a basis of $V$, it generates $V$, and $S$ is linearly independent. Applying EX 2 with the independent set $S$ and the generating set $\beta$ gives a subset $S_1$ of $\beta$ such that $S \cup S_1$ is a basis of $V$.

Proof. Let $c$ denote the set of all real sequences that converge in $\mathbb{R}$.

(It is a vector space.) $c$ is a subset of the space $\mathcal{F}(\mathbb{N}, \mathbb{R})$ of all real sequences. If $(a_n) \to a$ and $(b_n) \to b$, then by the limit laws $(a_n + b_n) \to a + b$, and for any scalar $k$, $(ka_n) \to ka$. Also the zero sequence converges to $0$. Hence $c$ is a subspace of $\mathcal{F}(\mathbb{N}, \mathbb{R})$, and so is itself a vector space over $\mathbb{R}$.

(It is infinite-dimensional.) For each $n \in \mathbb{N}$, define the sequence $e_n = (e_n(k))_{k=1}^\infty$ by $e_n(n) = 1$ and $e_n(k) = 0$ for $k \neq n$. All terms of $e_n$ except the $n$-th are $0$, so $e_n \to 0$, and in particular $e_n \in c$.

For each $n \in \mathbb{N}$, we show $\{e_1, \ldots, e_n\}$ is linearly independent. If $\sum_{i=1}^n c_i e_i = 0$ (the zero sequence), comparing the $k$-th term for $1 \le k \le n$ gives $\sum_{i=1}^n c_i e_i(k) = c_k = 0$. Hence $c_1 = \cdots = c_n = 0$, so $\{e_1, \ldots, e_n\}$ is linearly independent.

Now suppose $c$ were finite-dimensional, say $\dim c = d$. In a finite-dimensional vector space, every linearly independent set has at most $\dim c$ elements. But $\{e_1, \ldots, e_{d+1}\}$ is a linearly independent subset of $c$ with $d+1$ elements, a contradiction. Hence $c$ is infinite-dimensional.

Proof. Suppose $\dim_{\mathbb{Q}} \mathbb{R}$ is finite, and let $n = \dim_{\mathbb{Q}} \mathbb{R}$.

Two vector spaces over the same field with the same finite dimension are isomorphic, so $\mathbb{R}$ and $\mathbb{Q}^n$ are isomorphic as $\mathbb{Q}$-vector spaces; that is, there exists a bijective linear transformation $T : \mathbb{R} \to \mathbb{Q}^n$.

Since $\mathbb{Q}$ is countable, $\mathbb{Q}^n$, being a finite product of countable sets, is countable. Since $T$ is a bijection, $\mathbb{R}$ has the same cardinality as $\mathbb{Q}^n$, so $\mathbb{R}$ would also be countable. This contradicts the fact that $\mathbb{R}$ is uncountable (by Cantor's diagonal argument).

Hence $\dim_{\mathbb{Q}} \mathbb{R}$ cannot be finite, so $\mathbb{R}$ is an infinite-dimensional vector space over $\mathbb{Q}$.