Let $ (x_n) $ be a sequence of real numbers

and let $ L_{(x_n)} =\{l: \text{There exists a subsequence }(x_{n_j})\text{ of }(x_n)\text{ such that }\lim_{j} x_{n_j}=l \}$

Does there exist a sequence $ (x_n) $ such that

1) $ L_{(x_n)} =\Bbb Q $

2) $ L_{(x_n)} =\Bbb N $

3) $ L_{(x_n)}$ ={$1/n:n\in \Bbb N$}

My solution to 2) and 3) was that in 3) the infimum of $ L_{(x_n)}$ would be $0$ which is not included in $ L_{(x_n)}$ but $ \liminf x_n= \inf L_{(x_n)}$ then $0 \in L_{(x_n)}$ which is a contradiction. Similarly in 2) the sup of $ L_{(x_n)}$ would be infinity and the similar argument can be made. Is it correct? and what about the 1st one?

Any help would be appreciated Thanks

  • 1
    $\begingroup$ 2), yes. Enumerate the set $\{m+1/n : m,n\in \Bbb N, n\ne 0\}$. (Infinity is not a valid candidate for inclusion in $L_{(x_n)}$.) $\endgroup$ Jun 21, 2013 at 16:09

1 Answer 1


$L$ must be a closed subset of $\mathbb R$ (That's ok with (2) and what you essentially used in (3)). Given a seuqence $(y_n)$ in $L$ with $y_n\to y$ we find a subsequence by chosing $n_j$ such that $|x_{n_j}-y_j|<\frac 1n$ (and of course $n_j>n_{j-1}$). Then $x_{n_j}\to y$.

Applying this to (1) we see that $\mathbb Q\subseteq L$ implies $\overline{\mathbb Q}=\mathbb R\subseteq L$,so definitely $L\ne\mathbb Q$.

Likewise in (3), we have that $0\in\overline{\{1/n:n\in\mathbb N\}}$ and therefore $L$ cannto equal $\{1/n:n\in\mathbb N\}$.

In (2), the set $\mathbb N$ is closed, so the argument above does not show us that $L=\mathbb N$ is impossible. And indeed, $L=\mathbb N$ is possible. For example, let $x_n=n-\lfloor \sqrt n\rfloor^2$. You argue that $\infty$ is missing, but note that $\infty$ is not a limit of a sequence of real numbers (or can you, given $\epsilon>0$ find $N$ such that $x>N$ implies $|x_n-\infty|<\epsilon$?) Writing $\lim_{n\to\infty}x_n=+\infty$ is an abuse of notation to talk about a specific type of divergence.. On the other hand, one can work in the extended reals $\mathbb R\cup \{-\infty,+\infty\}$, endow them with a topology and then give sense to $\lim_{n\to\infty}=+\infty$. If that is your setup, then indeed $\mathbb N$ is not closed with respect to this topology, hence again we conclude that $L=\mathbb N$ is impossible.

  • $\begingroup$ Thanks a lot! but I have just started the real analysis unit and haven't topology yet. What we have been doing is say if a sequence diverges to infinity we had $L$= {$ \infty $} does that mean it is the $\mathbb R\cup \{-\infty,+\infty\}$ topology we were considering and then is $ L=\Bbb N $ impossible? $\endgroup$
    – user83369
    Jun 21, 2013 at 16:57

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