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

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.

• Yes, it looks just fine.
– Snaw
Commented Apr 18, 2022 at 20:35

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.$$\quad\blacksquare$$

• "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 $\lfloor x \rfloor$. Commented Apr 18, 2022 at 21:25
• 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 $w\notin (n_w,\infty)=K_{n_w}.$" Commented Apr 18, 2022 at 21:34
• 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! :) Commented Apr 18, 2022 at 21:39
• 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. Commented Apr 18, 2022 at 22:00
• Got it, thank you @Hanno! Commented Apr 18, 2022 at 22:13