Does it make sense to take the limit of an indexed set?

Question

Let $S_i= [0,1/i]$for instance, where $i\in \mathbb{N}$, would it make sense to say $\lim_{i \to \infty} S_i = [0,0] =\{0\}$ for instance, if not , why not?

Answer 1

I asked a related question before: Is the limit of a closed interval closed?

In the answer it is explained that an expression like '$\lim_{n\rightarrow\infty}S_i$' is only well defined if $\limsup_{i \rightarrow \infty} S_i = \liminf_{i \rightarrow \infty} S_i$. Then we may define $\lim_{n\rightarrow\infty}S_i := \limsup_{i \rightarrow \infty} S_i$. Since the sequence of your intervals is non-increasing, you can think of the limit as taking the infinite intersection (as was also mentioned in the comments). So $\lim_{n \rightarrow \infty} S_n = \bigcap_{n = 1}^{\infty} S_n$.

Answer 2

There is another interpretation. The sets $S_i$ are compact subsets of the metric space $\mathbb R$.

The set of all compact subsets of a metric space $(M,d)$ can be endowed with the Hausdorff metric. There are various equivalent definitions. Here is one:

For each subset $X \subset M$ and each $\epsilon > 0$ let $X_\epsilon = \bigcup_{x \in X} \{ z \in M \mid d(z, x) \le \epsilon \}$ which is sometimes called the $\epsilon$-fattening of $X$ or a generalized ball of radius $\epsilon$ around $X$. Then the Hausdorff distance between $X$ and $Y$ is defined as $d_{H}(X,Y) = \inf\{\epsilon \geq 0\mid X\subset Y_{\varepsilon }{\text{ and }}Y\subset X_{\varepsilon }\}$.

With the Hausdorff metric you easily see that $\lim_{i \to \infty} S_i = \{0\}$.