How to prove that for every natural number $k >= 1$, the following holds: $\bigcap_{n=1}^{k}(n, \infty) \neq \emptyset$ ?
I know that for every $k$, all the intervals will have the intersection: $(k, \infty)$
But how do I write this as a proof?
Asked
Active
Viewed 43 times
2
-
Well, certainly $(k,\infty) \neq \varnothing$ because $k+1\in(k,\infty)$. – MPW Mar 14 '16 at 12:25
-
2$$k+1 \in \bigcap_{n=1}^k (k, \infty )$$ – Crostul Mar 14 '16 at 12:26
1 Answers
3
You don’t need to prove that in order to answer the question: just show that $k+1$ is in the intersection. However, it’s not too hard to write down an actual proof that the intersection is $(k,\infty)$.
- First show that if $x>k$, then $x\in(n,\infty)$ for $n=1,\ldots,k$.
- Then show that if $x\le k$, then $x\notin(k,\infty)$, and therefore $x\notin\bigcap_{n=1}^k(n,\infty)$.
Brian M. Scott
- 616,228