1

I am pretty new with well-founded relations and the concept, so i have sort of gave a brief reason nothing to in depth would i be correct with what i have said any help and explanations are greatly appreciated.

This is what i have said i believe the relation to not be well-founded am i correct?

the order relation to < strictly less than on the positive rational numbers is not a well-founded relation as it has an infinite descending chain, for example this infinite descending chain ... $< a_n < ...< a_4 < a_3< a_2 <a_1 <a_0$ where $a_0 = \frac{9}{10}$ then ($... < an < ... <\frac{1}{5} < \frac{1}{4}< \frac{1}{3} <\frac{1}{2} <\frac{9}{10}$ ) this would be an infinite descending chain.

mark
  • 123

2 Answers2

1

It could be stated more clearly, but your reasoning is correct. There’s no need to use $0.99$; you could simply let $a_n=\frac1{n+1}$ for $n\in\Bbb N$ and observe that $\langle a_n:n\in\Bbb N\rangle$ is an infinite strictly descending chain in $\Bbb Q^+$. Another simple example is to let $a_n=2^{-n}$ for $n\in\Bbb N$.

(By the way, the usual term for $<$ is strictly less than.)

Brian M. Scott
  • 616,228
  • ah yeah i meant strictly less than my bad but thanks for your answer, also what is meant by this i haven't seen that before "⟨an:n∈N⟩" – mark Feb 11 '21 at 19:37
  • @James: You’re welcome. That’s one standard notation for the sequence $$\langle a_0,a_1,a_2,\ldots\rangle,.$$ – Brian M. Scott Feb 11 '21 at 19:38
  • got you your help is appreciated also what is meant by "Q+" that you have used is that the set notation for non-negative rational numbers? @Brian M. Scott – mark Feb 11 '21 at 19:40
  • @James: $\Bbb Q^+$ is one notation for the positive rationals; another that you might encounter is $\Bbb Q_{>0}$. For the non-negative rationals you might see $\Bbb Q_{\ge 0}$. – Brian M. Scott Feb 11 '21 at 19:41
  • great that clears everything up thanks for the help! – mark Feb 11 '21 at 19:43
  • @James: Excellent! You’re very welcome. – Brian M. Scott Feb 11 '21 at 19:44
0

A binary relation $R$ on a class $X$ is said to be well founded if every non empty subset $S \subseteq X$ has a minimal element.

Obviously, $\lt$ is not well founded on $\mathbb Q$ as $\mathbb Q$ itself has no minimal element: any $q$ can’t be a minimal element of $\mathbb Q$ as $q-1 \lt q$.

In your question, you used the infinite descending chain characterization of well founded that requires a form of axiom of choice.

mathcounterexamples.net
  • 70,018