2
$\begingroup$

I did a proof for exercise 2.5.9 of Introduction to Real Analysis: 4th edition by Bartle and Sherbert. The exercise is:

Let $K_n:=(n,\infty)$ for $n\in\mathbb{N}.$ Prove that $\bigcap_{n=1}^\infty K_n=\emptyset.$

My proof was significantly different than the solution manual, so I wanted to see if it's still correct, and if not, is there any way I can improve? Thank you!

Proof: Assume to the contrary that

$$ \bigcap_{n=1}^\infty K_n\neq\emptyset.$$

Then assume $x\in\bigcap_{n=1}^{\infty} K_n.$ Then we have $x>n$ for all $n\in\mathbb{N}$, which implies $x$ is an upper bound of $\mathbb{N}$. However, by the Archimedean Property*, $\mathbb{N}$ is not bounded above, so this is a contradiction.

$\therefore\bigcap_{n=1}^\infty K_n=\emptyset.\quad\blacksquare$

*Note that in the lectures for my Real Analysis class, the Archimedean Property was simply defined as "$\mathbb{N}$ is not bounded above," and we proved this fact in class.

$\endgroup$
1
  • 4
    $\begingroup$ Yes, it looks just fine. $\endgroup$
    – Snaw
    Commented Apr 18, 2022 at 20:35

1 Answer 1

2
$\begingroup$

Improvement is a relative entity, just let me offer the direct way to solve the exercise you set out.

You need not pass by contraposition, the direct approach is a more direct path towards $\blacksquare\,$:
Fix an arbitrary $x\in\mathbb R$.
Then choose $n\in\mathbb N$ such that $x\leqslant n$, whence $x\notin K_n=(n,\infty)\,$.
(E.g., $n=\lfloor x\rfloor +1$ does the job.)
Thus, the intersection of all $K_n$ is empty.$\quad\blacksquare$

$\endgroup$
5
  • $\begingroup$ "Then choose $n\in\mathbb N$ such that $x\leqslant n$". The existence of such $n$ is exactly the Archmedean property in the form mentioned by the OP. In this scheme, that is a more elementary thing than the existence of $\lfloor x \rfloor$. $\endgroup$
    – GEdgar
    Commented Apr 18, 2022 at 21:25
  • $\begingroup$ The solution manual's solution was really similar to this, actually (note this manual only gives partial solutions). It said "If $z\leq 0$, then $z\notin K_1$. If $w>0,$ then it follows from the Archimedean Property that there exists $n_w\in\mathbb{N}$ with $w\notin (n_w,\infty)=K_{n_w}.$" $\endgroup$ Commented Apr 18, 2022 at 21:34
  • $\begingroup$ And as @GEdgar mentioned, we haven't talked about the floor function yet, so I'm not sure if I'd be allowed to include that. Maybe if I omitted the e.g. part in parentheses I could use your proof too. Thanks for your help! :) $\endgroup$ Commented Apr 18, 2022 at 21:39
  • 1
    $\begingroup$ You are welcome @blakedylanmusic . That was precisely the intention of putting the parentheses, i.e, thatits contents may be omitted. The focus of my two cents is on the directness, not on the Archimedean property. $\endgroup$
    – Hanno
    Commented Apr 18, 2022 at 22:00
  • $\begingroup$ Got it, thank you @Hanno! $\endgroup$ Commented Apr 18, 2022 at 22:13

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .