The set {${{x_i} : i ≥ 1}$} has volume zero.

Question

Let be {${x_i}$} a sequence in $\mathbb R^n$ that converges to $x$. Then the set {${{x_i} : i ≥ 1}$} has volume zero.

Can someone explain to me why is this true?

Answer 1

For $\varepsilon > 0$, there are at most finitely many terms of the sequence/elements of the set that are outside a ball with volume $\varepsilon$ centered at $x$. Those terms can be covered with finitely many balls with volumes of the form $\varepsilon/2^k$ for distinct values of $k>0$, hence with total volume less than $\varepsilon$. Hence, for all $\varepsilon > 0$, the set can be covered with balls of total volume less than $2\varepsilon$.

More generally, any set in $\mathbb{R}^n$ that is at most countable has volume zero: for every $\varepsilon >0$, the elements of the set can be covered with balls of volumes of the form $\varepsilon/2^k$ for distinct $k > 0$, hence with total volume at most $\varepsilon$.

Answer 2

Intuitively, it should be easy to guess that the set $S=\{x_i : i \geq 1\}$

After all, if we take the special case $n=3$ and $x_i=\begin{bmatrix} 0 \\ 0 \\ 1/i \end{bmatrix} $, then $S$ will be a subset of the z-axis. Since $S$ will be the subset of a line, and the line obviously has volume zero (atleast when we work in $\mathbb{R^3}$), $S$ will have volume zero also.