Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

If I show this statement:

$$x\in \left] a,b \right[ \Rightarrow \exists n \in \mathbb{N} : x\in \left] -\frac1{n}, 1+\frac1{n}\right[$$

Have I then shown this statement:

$$]a,b[ \subseteq \bigcup_{n=1}^\infty \left] -\frac1{n}, 1+\frac1{n}\right[\qquad ?$$

share|cite|improve this question
Yes. And conversely if you have proved the second, you have proved the first. Whether the two statements are "the same" is a more complicated matter. It all depends on the exact definitions that have been given for $\subseteq$, for $\cup_{n=1}^\infty$, and so on. The two statements are likely not the same, though they certainly are equivalent. – André Nicolas Oct 4 '11 at 15:45
What do the opposite brackets mean? – mixedmath Oct 5 '11 at 1:58
up vote 1 down vote accepted

Note that:

  • A set $A$ is a subset of a set $B$ if and only if $x\in A\Rightarrow x\in B$.

  • If we have $A_i$ then $x\in\bigcup A_i$ if and only if for some $i$ we have $x\in A_i$.

Combine the two result and you have indeed what you wanted.

Note, while at it, that $(-1,2)$ which is the interval for $n=1$ (since $-\frac{1}{1}=-1,\ 1+\frac{1}{1}=2$) is a superset of all the other intervals. In particular this whole union is just $(-1,2)$ to begin with.

share|cite|improve this answer
Actually, as the supremum of each interval is $1+1/n$, not $1/n$, the entire union is $(-1, 2)$, not $(-1, 1)$. – jwodder Oct 4 '11 at 16:59
@jwodder: Thanks for the correction. – Asaf Karagila Oct 4 '11 at 17:39

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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