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3 logic
2 Clarified what was meant by "constructive"

I am searching for a constructive proof of the following fact: If $X$ is an infinite set, there exists an uncountable family `$(X_\alpha)_{\alpha \in A}$` of infinite subsets of $X$ such that $X_\alpha \cap X_\beta$ is finite whenever $\alpha \neq \beta$. The way I know how to prove this statement is as follows.

First, it suffices to prove the case when $X$ is countable. Thus we can choose a bijection between $X$ and $\mathbb{Q} \cap [0,1]$. To save notation we can tacitly assume that $X = \mathbb{Q} \cap [0,1]$.

Let the index set be $A = [0,1] \setminus X$, i.e. all the irrationals in $[0,1]$. For each $\alpha \in A$, choose a sequence $(x_{\alpha 1},x_{\alpha 2},\dots)$ of elements of $X$ such that $x_{\alpha n} \to \alpha$ as $n \to \infty$, and let $X_\alpha = \{ x_{\alpha_n} \mid n \in \mathbb{N} \}$.

Since $\alpha$ is irrational, the sequence $(x_{\alpha n})$ cannot be eventually constant, so $X_\alpha$ is infinite. And if $\alpha \neq \beta$ then the sequences $(x_{\alpha n})$ and $(x_{\beta n})$ can have only finitely many terms in common since they have different limits, so $X_\alpha \cap X_\beta$ is finite.

Is it possible to do this in a more constructive way? I know very little about set theory and logic, so I apologize if this question is too elementary. Also, I wasn't sure about any relevant tags other than set-theory, so please feel free to add appropriate tags.

Edit: to clarify, I didn't have a clear notion of what I meant by "constructive" here. What I didn't like about the proof I gave above was that it required a choice of sequence of rationals converging to each irrational. The answers so far all address this concern adequately.

1

# Uncountable family of infinite subsets with pairwise finite intersections

I am searching for a constructive proof of the following fact: If $X$ is an infinite set, there exists an uncountable family `$(X_\alpha)_{\alpha \in A}$` of infinite subsets of $X$ such that $X_\alpha \cap X_\beta$ is finite whenever $\alpha \neq \beta$. The way I know how to prove this statement is as follows.

First, it suffices to prove the case when $X$ is countable. Thus we can choose a bijection between $X$ and $\mathbb{Q} \cap [0,1]$. To save notation we can tacitly assume that $X = \mathbb{Q} \cap [0,1]$.

Let the index set be $A = [0,1] \setminus X$, i.e. all the irrationals in $[0,1]$. For each $\alpha \in A$, choose a sequence $(x_{\alpha 1},x_{\alpha 2},\dots)$ of elements of $X$ such that $x_{\alpha n} \to \alpha$ as $n \to \infty$, and let $X_\alpha = \{ x_{\alpha_n} \mid n \in \mathbb{N} \}$.

Since $\alpha$ is irrational, the sequence $(x_{\alpha n})$ cannot be eventually constant, so $X_\alpha$ is infinite. And if $\alpha \neq \beta$ then the sequences $(x_{\alpha n})$ and $(x_{\beta n})$ can have only finitely many terms in common since they have different limits, so $X_\alpha \cap X_\beta$ is finite.

Is it possible to do this in a more constructive way? I know very little about set theory and logic, so I apologize if this question is too elementary. Also, I wasn't sure about any relevant tags other than set-theory, so please feel free to add appropriate tags.
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