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I have the following question: for an uncountably infinite set, $X$, will the following process ever terminate?

  1. Construct a sequence $\{x_{n}\}_{n\in \mathbb{N}}$ by setting $x_{n}$ equal to a distinct element of $X$, for each $n$;

  2. Construct another sequence, $\{y_{n}\}_{n \in \mathbb{N}}$, by setting $y_{n}$ equal to a distinct element of $X \setminus \{x_{n}\}_{n\ \in \mathbb{N}}$, for each $n$.

  3. Construct another sequence...

...

That is, will the set of elements of $X$ that are not included in any sequence, eventually be finite? All I know at this point is that termination of the process is consistent with $X$ being uncountably infinite (as all the latter means is that any sequence $\{z_{n}\}_{z\in \mathbb{N}}$ will always omit at least one element of $X$).

Charles
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    I am not sure of what your last paragraph meant, but if it meant you were already able to prove that the process will never terminate, then with barely more work you were also able to answer your question. – Anne Bauval Mar 09 '23 at 11:13

2 Answers2

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A sequence of sequences contains only countably many elements (see this question, for example), so the process can't exhaust an uncountable set.

Hans Lundmark
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Let $X$ be an uncountable set and $(x_{n})$ be a sequence of distinct members of $X$. $X $ is uncountable and $\{x_n\}$ is countable so we can prove that $X \backslash \{x_n\}$ is also uncountable as follows:
Say $X \backslash \{x_n\}$ were countable. $X \backslash \{x_n\} {\displaystyle \cup } \{x_n\} = X$ is a countable union of countable sets, and hence countable. This is a contradiction.
Then by the same arguement, since $X \backslash \{x_n\}$ is uncountable, $X \backslash \{ \{x_n\} {\displaystyle \cup } \{y_n\} \}$ is also uncountable and so on...
Inductively, we see that no matter how many sequences we exclude, we are left with an uncountable, and hence infinite, set.

When we see a statement like this, its often helpful to consider the implications. Even without this arguement, intuitively we can understand that its very unlikely that this process ever ends. If this process were to end, it would need to do so after a finite number of iterations, say $N$. Then what would $N$ be? $100? \ 1000? \ 10^{100000}?$ It just seems strange for such an $N$ to exist.

Ogglie Ostrich
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  • This does not answer the question, which was about "the set of elements of $X$ that are not included in any sequence", because you only explained ("inductively") that after excluding any finite number of sequences, there remains an uncountable set. – Anne Bauval Mar 09 '23 at 11:08
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    For the process to eventually "end" after excluding a certain amount of sets, the amount of sets we exclude has to be finite. I assumed this is what the question was asking. The process is defined by removing one sequence at a time and waiting for X to be finite, not removing an infinite union of sets. If we were to remove a countably infinite union of sets, X will still be countable, whereas if we were to remove an uncountable union, we could have X countable, uncountable or finite depending on our method. Since the latter interpretation isn't a well defined problem, I assumed the former. – Ogglie Ostrich Mar 09 '23 at 19:06