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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

Omojola Micheal
  • 4,504

1 Answers1

2

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

PROOF

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.

EXAMPLE 2

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}

PROOF

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

EXAMPLE 3

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.

EXAMPLE 4

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.

Omojola Micheal
  • 4,504