Infinite vector problem related to the continuum hypothesis

Question

I developed an algorithm that might map the countable ordinal numbers (ones with just a countable number of predecessors) into a subset of the reals without using the axiom of choice. It leads to the following question: Consider infinite dimensional vectors with only Natural numbers as entries. Comparison with continued fractions maps these vectors into the irrational numbers greater than 1. The cardinality of such vectors is the same as the cardinality of the real numbers. Consider any subset $V$ of those vectors and the set $W(V)$ which consists of all the finite sets that can be written as $\{(v_a,n_a)|v_a\in V, n_a \text{is a natural number, no repeats among the }v_a\}$. For any $w \in W(V)$ we define $sum(w) = \sum v_a n_a$. We wish to consider only $V$ such that $sum(w_\alpha) = sum(w_\beta)$ implies that $w_\alpha = w_\beta$. Essentially, $V$ is a basis for $\{sum(w)|w \in W(V)\}$. Is there an uncountable set $V$ with those properties? If not the output of the algorithm is only a countable set. If so the algorithm probably defines an uncountable subset of the real numbers that is well ordered. I would prefer a proof that such a $V$ exists without using the axiom of choice. Off hand if it wasn't for the restriction that the sums are all finite, the answer would clearly be no.

Answer 1

Yes, there is such uncountable set.

Note that if we can find an uncountable family $X$ of infinite subsets of $\mathbb N$ s.t. intersection of any two sets from the family is finite, we can build necessary $V$ from it: for each $\alpha \in X$ add $v_\alpha$ to $V$ - vector that has $1$ on coordinates with numbers in $\alpha$, and $0$ on others.

Assume that we have $\sum_\limits{\alpha \in Y} n_\alpha v_\alpha = 0$ for some finite $Y \subset X$ and non-zero integers $n_\alpha$. Fix some $\alpha_0 \in Y$. By property of $X$, $\alpha_0 \cap \left(\bigcup\limits_{\alpha \in Y, \alpha \neq \alpha_0} \alpha\right)$ is finite, and thus $\alpha_0 \setminus \left(\bigcup\limits_{\alpha \in Y, \alpha \neq \alpha_0} \alpha\right)$ is non-empty. Choose some $i$ from it - then $i$-th coordinate of our sum is equal to $n_{\alpha_0}$, and, as the sum is $0$, $n_{\alpha_0} = 0$.

To construct such $X$, see, for example, this answer.

However, it's not clear, how you are going to construct uncountable well-ordered subset of $\mathbb R$ using such $V$. You definitely can't build an uncountable well-ordered subset of $\mathbb R$ in ZF, because axiom of determinacy is consistent with ZF, and from it you have that 1) any uncountable subset of $\mathbb R$ has cardinality of continuum; 2) continuum can't be well-ordered.