When I initially moved from Applied Mathematics background to Pure Mathematics (Graduate school), things were very tough in terms of proofs but they have gotten much better now. At those times, I couldn't explain what the following meant in proofs;

  1. Let $\epsilon>0$ be given;
  2. Let $n\in\Bbb{N}$ be fixed;
  3. Let $x\in\Bbb{R}$ be fixed but arbitrary;
  4. Take $n=N+1,$ for some $N\in\Bbb{N}$;
  5. In particular, for $\epsilon=1/2$;
  6. Let $\epsilon'>0$, for $\epsilon=\epsilon'/3$
  7. Such that;
  8. Whenever,
  9. For each;
  10. For all, and many more.

Could you please, explain these in details, perhaps, with some examples? That is, when to use them and how to... If you have other examples, I'd be glad to learn from you. Thanks

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    $\begingroup$ Related: When do I use “arbitrary” and/or “fixed” in a proof?. $\endgroup$ – StackTD Jan 4 at 15:00
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    $\begingroup$ These four statements could be replaced with: 1. Let $\epsilon > 0$. 2. Let $n \in \mathbb N$. 3. Let $x \in \mathbb R$. 4. Let $n = N + 1$. $\endgroup$ – littleO Jan 4 at 15:09
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    $\begingroup$ These fixed terms are somewhat arbitrary ... $\endgroup$ – Hagen von Eitzen Jan 4 at 15:40
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    $\begingroup$ "In particular, for $x=3$: we're going to temporarily let $x=3$ to examine that case, but $x$ is still a variable and not defined to be always $3$. $\endgroup$ – timtfj Jan 4 at 15:44
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    $\begingroup$ "Fixed but arbitrary" we can choose any value we like, but once chisen, it's a constant. $\endgroup$ – timtfj Jan 4 at 15:48

1. Let $\epsilon>0$ be given; Take for example, Prove that $1/n\to 0,$ as $n\to\infty.$


Let $\epsilon>0$ be given (it means give me a specific positive nuimber, say $\epsilon$, then I'd prove to you that$ \;\forall\; \epsilon>0$, $1/n<\epsilon$ for large n). Hence, given any $\epsilon>0$, choose $N=\left[1/\epsilon+1\right]$, then, \begin{align} \dfrac{1}{n}\leq \dfrac{1}{N}<\epsilon,\;\forall\; n\geq N. \end{align} This implies that for all $\epsilon>0$, \begin{align} \dfrac{1}{n}<\epsilon,\;\forall\; n\geq N. \end{align}

2. Let $n\in\Bbb{N}$ be fixed;

Pick a particular $n\in\Bbb{N}$ for the whole proof which cannot be changed in the course of proving.


Prove that for bounded positive real sequences $\{a_n\}_{n\in\Bbb{N}}$ \begin{align} \limsup_{n\to\infty} \sqrt[n]{|a_n|}\leq\limsup_{n\to\infty} \left|\dfrac{a_{n+1}}{a_n} \right| \end{align}


Let $\epsilon>0$ be given and $\beta:=\limsup_{n\to\infty} \left|\dfrac{a_{n+1}}{a_n} \right| $, then there exists $N$ such that \begin{align} \left|\dfrac{a_{n+1}}{a_n} \right|<\beta+\epsilon,\;\;\forall\;n\geq N. \end{align} Let $n\geq N$ be fixed, \begin{align} \left|\dfrac{a_{n}}{a_N} \right|=\left|\dfrac{a_{n}}{a_{n-1}} \right|\cdot \left|\dfrac{a_{n-1}}{a_{n-2}} \right|\cdots\left|\dfrac{a_{N+1}}{a_{N}} \right|<\left(\beta+\epsilon\right)^{n-N}\end{align} This implies \begin{align} \sqrt[n]{\left|a_{n}\right|}<\sqrt[n]{\left|a_N \right|}\left(\beta+\epsilon\right)^{1-N/n}\end{align} Taking $\limsup$, \begin{align} \limsup_{n\to\infty} \sqrt[n]{|a_n|}\leq\beta+\epsilon\end{align} Since $\epsilon>0$ was arbitrary, \begin{align} \limsup_{n\to\infty} \sqrt[n]{|a_n|}\leq\limsup_{n\to\infty} \left|\dfrac{a_{n+1}}{a_n} \right| \end{align}

3. Such that, arbitrary


Prove that $f:X\to\overline{\Bbb{R}}$ is lower semi-continuous if

\begin{align} \forall\;\lambda\in \Bbb{R},\;\;f^{-1}\left((\lambda,\infty] \right)\;\;\text{is open in}\;X. \end{align} PROOF

Let $\lambda\in \Bbb{R}$ and $x_0\in X$ be arbitrary, such that $\lambda<f(x_0)$. Then, \begin{align} x_0\in f^{-1}\left((\lambda,\infty] \right). \end{align} Take \begin{align} V=f^{-1}\left((\lambda,\infty] \right), \;\;\text{where }\;\;V\in U\left(x_0 \right).\end{align} Let $x\in V$, then $f(x)\in (\lambda,\infty] ,$ which implies that $f(x)>\lambda.$ Hence, for all $\lambda\in \Bbb{R}$ and $x_0\in X$ such that $\lambda<f(x_0)$, $f(x)>\lambda,\;\forall\;x\in V.$ This implies that that $f:X\to\overline{\Bbb{R}}$ is lower semi-continuous.

4. Fixed but arbitrary, fixed, such that and whenever.


Prove that $|x|$ is continuous on $\Bbb{R}$.

Let $\epsilon>0$ be given, $x\in \Bbb{R}$ be fixed but arbitrary and $x_0\in \Bbb{R}$ be fixed, such that $|x-x_0|<\delta,$ then \begin{align} \left| |x|-|x_0| \right| \leq \left| x-x_0 \right|<\delta.\end{align} So, given any $\epsilon>0$, choose $\delta=\epsilon,$ then \begin{align} \left| |x|-|x_0| \right|<\epsilon,\;\;\textbf{whenever}\;\;\left| x-x_0 \right|<\delta\end{align} I pick any $x\in\Bbb{R}$ for the proof but it will be fixed throughout the proof.

  • $\begingroup$ Do you know of cases where given (from point 1), fixed (from point 2), and fixed but arbitrary (from point 3) are NOT interchangeable? $\endgroup$ – Todor Markov Jan 4 at 16:37
  • $\begingroup$ @Todor Markov: They, can be but I was trying to find other words of expressing them $\endgroup$ – Omojola Micheal Jan 4 at 19:15

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