2
$\begingroup$

Intersection of different sets mean that we will get only the elements that exist in each of them.

Then why intersection of all $\mathbb{Z}^+$ numbers will yield $\{1\}$?

It is clear that $1$ only exists once in the collection, and $2,3,4,\dots$ none of them match $1$, if we say we don't have duplicates.

$$ \bigcap_{i=1}^\infty A_i = \{1\}. $$ $$ \bigcap_{i=1}^\infty \{1,2,3,\ldots,i\} = \{1\}. $$

$\endgroup$
12
  • $\begingroup$ Intersection is defined as an operation between sets. which sets exactly are you referring to by "all +Z numbers"? $\endgroup$ – barak manos Apr 12 '16 at 4:51
  • $\begingroup$ @barakmanos ∞ (intersection of collection)i=1 {1,2,3,...,i}={ 1}. $\endgroup$ – Rafsan Mobasher Apr 12 '16 at 4:51
  • $\begingroup$ I could ask what you mean by an infinite integer. The term is meaningful in some contexts, and may have different meanings in different contexts. However, I have notice some people saying things like "infinite integers" when they mean "infinitely many integers". The former is an incorrect usage. $\qquad$ $\endgroup$ – Michael Hardy Apr 12 '16 at 4:52
  • $\begingroup$ I formatted some of your post. Can't really figure out what you mean by "∞ (big intersection) i=1". $\endgroup$ – barak manos Apr 12 '16 at 4:53
  • $\begingroup$ @MichaelHardy Infinitely many integers. $\endgroup$ – Rafsan Mobasher Apr 12 '16 at 4:53
3
$\begingroup$

\begin{align} & \{1\} \\ & \{1,2\} \\ & \{1,2,3\} \\ & \{1,2,3,4\} \\ & \{1,2,3,4,5\} \\ & \qquad \vdots \end{align}

The only thing that is a member of all of the sets above is $1$.

$\endgroup$
2
  • $\begingroup$ +1, but I think that "only" should also be in bold. $\endgroup$ – barak manos Apr 12 '16 at 5:01
  • $\begingroup$ I see.I thought each number is representing numbers, but they are representing collection of numbers starting from 1.Many thanks. $\endgroup$ – Rafsan Mobasher Apr 12 '16 at 5:01

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.