This is in the context of "set theory" and "sequence of sets" and their corresponding notations.
I was trying to solve a question which says:
- $C_1, C_2, C_3 \dots$ is a non-decreasing sequence of sets, i.e., ${C_k} \subset C_{k+1}\space$for $k = 1, 2, 3\dots$ ; and
- $\lim \limits_{k\space\to\space\infty}C_k$ is defined as the union $\space C_1 \cup C_2 \cup C_3\dots\space$ = $\bigcup_ \limits{n\space=\space1}^\infty C_n\space$for $n = 1, 2, 3 \dots\space$.
What is $\lim \limits_{k\space\to\space\infty}C_k$ where $C_k = \big\{x:\frac{1}{k} \le x \le 4-\frac{1}{k^2}\big\}\space$for $k = 1, 2, 3 \dots\space$?
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I approached the problem as follows:
Since $C_k = \big\{x:\frac{1}{k} \le x \le 4-\frac{1}{k^2}\big\}\space$for $k = 1, 2, 3 \dots\space$; we have,
$C_1 = \big\{x:1 \le x \le 3\big\}\space \implies \space x\in \left[1,\space3\right]$ for $C_1$,
$C_2 = \big\{x:\frac{1}{4} \le x \le \frac{15}{4}\big\}\space \implies \space x\in \left[\frac{1}{4},\space\frac{15}{4}\right]$ for $C_2$,
and so on.
Hence,
$\begin{align}
\lim \limits_{k\space\to\infty}C_k &= \bigg\{x:\bigg(\lim \limits_{k\space\to\space\infty}\frac{1}{k}\bigg) \le x \le \bigg(\lim \limits_{k\space\to\infty}\big(4-\frac{1}{k^2}\big)\bigg)\bigg\} \\
&=\bigg\{x:\epsilon \le x \le 4-\delta\bigg\}
\end{align}$
where $\space\epsilon\to0$, $\space\epsilon > 0$,
and $\space\delta\to0$, $\space\delta > 0$.
So, for $\lim \limits_{k\space\to\space\infty}C_k\space$,$\space$ we have, $\space$$x\in \left[\epsilon,\space4-\delta\right]\space$ where $\epsilon$ and $\delta$ are infinitesimals.
The interval $\left[\epsilon,\space4-\delta\right]$ is better represented by the interval $\left(0,\space4\right)$.
Hence the answer is $\lim \limits_{k\space\to\space\infty}C_k = x$ where $x \in (0,\space4)$.
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My questions are:
- Is my approach correct? Is there a different way to understand and solve this question?
- Are my notations correct for the lines involving the "infinitesimals" along with the "belongs to" and the "less than or equal to" symbols for sets and/or intervals (like the ones below)?
- $\bigg\{x:\bigg(\lim \limits_{k\space\to\space\infty}\frac{1}{k}\bigg) \le x \le \bigg(\lim \limits_{k\space\to\infty}\big(4-\frac{1}{k^2}\big)\bigg)\bigg\}$
- $\bigg\{x:\epsilon \le x \le 4-\delta\bigg\}$
- $x\in \left[\epsilon,\space4-\delta\right]$ where $\epsilon$ and $\delta$ are infinitesimals.
- The interval $\left[\epsilon,\space4-\delta\right]$ is better represented by the interval $\left(0,\space4\right)$.
Best Answer
I would completely avoid a discussion of infinitesimals. Much more simply, an element $x$ is in the union of the $C_k$ iff. it is in one $C_k$ for some $k$. If $x\in(0,4)$, then there are large enough $k$ that $1/k\le x\le4-1/k^2$. So $(0,4)\subseteq\lim C_k$.
If $x\le 0$, then for no finite positive $k$ can we have $1/k\le x$. Likewise, if $x\ge4$ then $x$ cannot be in any of the individual sets $C_k$ and therefore is not in their union. So $(0,4)=\lim C_k$.
It is not right to say $(0,4)=[\epsilon,4-\delta]$ for some “infinitesimals”. Infinitesimals are really rather tricky to rigorously define, and are generally not needed. They are certainly not needed here, and I doubt their use is even correct here. After all, $[\cdots]$ shall be a closed interval but $(\cdots)$ an open one.
Notions of “limit of a set” must not be conflated with the limits in real analysis or general topology. Rather they are conveying a similar idea: that does not mean your approach to understand them must also involve $\epsilon,\delta$. This concept of a limit of a sequence of sets can be applied in any general set: real numbers need not apply.