show/hide this revision's text 3 revised construction

The answer is no, I think. Here is a proof sketch. (with one An unclear point in a previous version has now been removed, for now.by slightly modifying the construction of the sequence.)

[Revised version from here on.]

I claim that $W$ has no will show

  • From any basis . So assume that $C$ is of $W$ we can define a basis.

    Fix 1-1 sequence of elements of $W$.

  • From any well-order 1-1 sequence of the finite subsets elements of $\omega$. Take the first W$ we can define a 1-1 sequence of elements of $S$. Together, this will show that there is no basis, as $S$ contains no countably infinite set.

    • For each set $D_0$ D$ which appears as the domain of a basis vector. Add , let $x_D$ be the sum of all basis vectors with domain $D_0$; the this domainof their sum . So $s_0$ is non-empty x_D \neq 0$, and has a least element $n_0$; now for $s_0$ chooses one element D\neq D'$ we get $x_0 \in S_{n_0}$x_D\neq x_{D'}$. (Namely, the one with coordinate 1$.)

    Let $D_1$ be From a well-order of the next set which appears as domain finite subsets of $\omega$ we thus obtain a basis vectorwell-ordered sequence of nonzero vectors. Continue as above to find $s_1$, and let $n_1$ Since there must be the least element in infinitely many basis vectors, and only finitely many can share the domain same set $D$, we have obtained an infinite sequence of vectors in $s_1$ other than W$.

    We are now given an infinite sequence $n_0$, and let (w_n)$ of distinct vectors of $s_1$ choose an element W$. The union of their domains cannot be finite, so we may wlog assume that the sequence $x_1 k_n:= \in S_{n_1}$max(dom(w_n))$ is strictly increasing. (It may happen that Thin out, if necessary.)

    Now let $a_n$ be the domain element of $s_1$ is the singleton S_{k_n}$ be such that $\{n_0\}$; in this case, c_{a_n}(w_n)=1$. Then the set of those $x_1$ is undefined.)

    Continue by induction, and check that a_n$ meets infinitely many of the $x_k$ will be definedS_k$ in a singleton. (Not quite sure about that)
    show/hide this revision's text 2 backslashes. point out unclear point.

    The answer is no, I think. Here is a proof sketch (with one unclear point, for now.)

    Let $(S_n)_{n\in\omega}$ be a family of ``pairs of socks''; that is, each $S_n$ has 2 elements, the $S_n$ are disjoint, but there is no set which meets infinitely many $S_n$ in exactly one point. Let $S$ be the union of the $S_n$.

    Let $V$ be a vector space with basis $S$ over the 3-element field. For each $v\in V$, each $s\in S$ let $c_s(v)$ be the $s$-coordinate of $v$. (In your notation: $v(s)$.)

    Consider the subspace $W$ of all vectors $w$ with the following property: For all $n$, if $S_n = \{a,b\}$, then $c_a(w)+c_b(w)=0$. The set of all $n$ such that for both/any $a\in S_n$ we have $c_a(w) \not=0$ neq0$ will be called the domain of $w$. Clearly, each domain is finite, and for each finite subset of $\omega$ of size $k$ there are $2^k$ vectors $w\in W$ with this domain.

    I claim that $W$ has no basis. So assume that $C$ is a basis.

    Fix any well-order of the finite subsets of $\omega$. Take the first set $D_0$ which appears as the domain of a basis vector. Add all basis vectors with domain $D_0$; the domain of their sum $s_0$ is non-empty and has a least element $n_0$; now $s_0$ chooses one element $x_0 \in S_{n_0}$. (Namely, the one with coordinate 1$.)

    Let $D_1$ be the next set which appears as domain of a basis vector. Continue as above to find $s_1$, and let $n_1$ be the least element in the domain of $s_1$ other than $n_0$, and let $s_1$ choose an element $x_1 \in S_{n_1}$. (It may happen that the domain of $s_1$ is the singleton $\{n_0\}$; in this case, $x_1$ is undefined.)

    Continue by induction, and check that infinitely many $x_k$ will be defined. (Not quite sure about that)
    show/hide this revision's text 1

    The answer is no, I think. Here is a proof sketch (with one unclear point, for now.)

    Let $(S_n)_{n\in\omega}$ be a family of ``pairs of socks''; that is, each $S_n$ has 2 elements, the $S_n$ are disjoint, but there is no set which meets infinitely many $S_n$ in exactly one point. Let $S$ be the union of the $S_n$.

    Let $V$ be a vector space with basis $S$ over the 3-element field. For each $v\in V$, each $s\in S$ let $c_s(v)$ be the $s$-coordinate of $v$. (In your notation: $v(s)$.)

    Consider the subspace $W$ of all vectors $w$ with the following property: For all $n$, if $S_n = \{a,b\}$, then $c_a(w)+c_b(w)=0$. The set of all $n$ such that for both/any $a\in S_n$ we have $c_a(w) \not=0$ will be called the domain of $w$. Clearly, each domain is finite, and for each finite subset of $\omega$ of size $k$ there are $2^k$ vectors $w\in W$ with this domain.

    I claim that $W$ has no basis. So assume that $C$ is a basis.

    Fix any well-order of the finite subsets of $\omega$. Take the first set $D_0$ which appears as the domain of a basis vector. Add all basis vectors with domain $D_0$; the domain of their sum $s_0$ is non-empty and has a least element $n_0$; now $s_0$ chooses one element $x_0 in S_{n_0}$. (Namely, the one with coordinate 1$.)

    Let $D_1$ be the next set which appears as domain of a basis vector. Continue as above to find $s_1$, and let $n_1$ be the least element in the domain of $s_1$ other than $n_0$, and let $s_1$ choose an element $x_1 in S_{n_1}$. (It may happen that the domain of $s_1$ is the singleton $\{n_0\}$; in this case, $x_1$ is undefined.)

    Continue by induction, and check that infinitely many $x_k$ will be defined.