Is there any sequence coverall the real numbers? Like ∀x∈R ,∃n∈N s.t. a_n = x.
I think the answer is not, but I need help to prove this, if you have some argument please comment it.
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No, otherwise the real numbers would be countable. (Cf. Cantor‘s second diagonal argument.) – Maximilian Janisch Sep 24 '21 at 13:32
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See e.g. https://en.wikipedia.org/wiki/Cantor%27s_diagonal_argument#Real_numbers – Sep 24 '21 at 13:34
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To add to the previous comments: Think of it like this. The maximum cardinality of the image of the sequence is achieved if it's one to one. But then you have a bijection with the natural numbers. – ProfOak Sep 24 '21 at 13:41