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Let us consider the set $X$ of the following infinite dimensional vectors:

$$ x = (x_{1}, x_{2}, \dots, x_{k}, 0, \dots,), $$ such that $x_{i} \in \mathbb{Z}$ and for any $x$ there exist $k < \infty$ such that $x_{j} = 0$ for all $j > k$.

The statement: $X$ is not countable.

Try: I tried to show this using the fact that the set of all subsets of all integers is uncountable

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Your statement is wrong: The set is countable. Take a look at this question. That question assumes natural numbers, but the argument is the same for integers (or any countable set).

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  • $\begingroup$ awesome! I see now! yes, $X$ can be any countable set. $\endgroup$ – ABK Oct 29 '19 at 16:03

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