Limit of sequence of sets $S_n = \{\frac{1}{n}\}$

Question

Suppose I have a sequence of sets $$S_n = \left\{\frac{1}{n}\right\}$$ And I want to know $\displaystyle\lim_{n\to\infty} S_n$.

Intuition tells me that as with usual limits $\displaystyle\lim_{n\to\infty} S_n = \left\{0\right\}$. But by definition from this answer $\displaystyle\lim_{n\to\infty} S_n = S$ iff $$\forall x\; \exists N\in\mathbb N\; \forall n>N : x\in S\Leftrightarrow x \in S_n$$

Obviously $0$ will not appear in any $S_n$, when $n$ is finite.

Does that mean that $\displaystyle\lim_{n\to\infty} S_n$ doesn't exist? And if yes, then is there some other type/definition of limit that will make $S_n$ converge to $\{0\}$?

According to the linked definition the limit is the empty set. But you really should check the definition used in your book (or wherever this question came from). — Jyrki Lahtonen, May 18 '20 at 05:31

score 1 · Accepted Answer · answered May 18 '20 at 05:37

1

In the set theoretic sense the limit of any disjoint sequence of sets is empty. However $S_n$ does converge to $\{0\}$ w.r.t the Hausdorff metric. Ref. https://en.wikipedia.org/wiki/Hausdorff_distance

answered May 18 '20 at 05:37

Kavi Rama Murthy

311,013

Thanks. I think Hausdorff convergence is what I was looking for – emptysamurai May 18 '20 at 05:45

Limit of sequence of sets $S_n = \{\frac{1}{n}\}$

1 Answers1