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Let $(u_{n})$ and $(v_{n})$ be two convergente sequences to the same limit $1$. We have $v_{n}>u_{n}$, then there exist infinitely many reals $a$ such that $u_{n}<a<v_{n}$ for all natural number $n$. Hence the set of those $a$ is an uncountable set. When $n→∞$, we must get $a=1$.

So, my question is: How I can explain this contradiction regarding the number of elements in the set of $a$'s.

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It is not true that there are infinitely many reals $a$ such that $u_n<a<v_n$ for all natural numbers $n$. What is true is that for each natural number $n$ there are infinitely many reals $a$ such that $u_n<a<v_n$ for that $n$. In fact for any real number $a\ne 1$ there is an $m\in\Bbb N$ such that for all $n\ge m$, $a\notin(u_n,v_n)$; that is, each real number different from $1$ is in only finitely many of the intervals $(u_n,v_n)$. (You should try to prove this; it’s a fairly simple consequence of the definition of convergence to a limit.)

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