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

Suppose you have a doubly indexed sequence of reals, $(\alpha_{ij})$. Why is $$ \sup_i \;\sup_j\ \alpha_{ij}=\sup_j\;\sup_i\ \alpha_{ij}? $$ I know one approach is to note $\alpha_{mn}\leq\sup_j\;\sup_i\alpha_{ij}$ for any $m$ and $n$. Why is this exactly? I don't really know what $\sup_j\sup_i\alpha_{ij}$ means. Does it mean first fix some $j$ and find the supremum of $\alpha_{ij}$ as $j$ remains fixed as $i$ runs over $\mathbb{N}$? And then after that, find the supremum of $\sup_i\alpha_{ij}$ as $j$ runs over $\mathbb{N}$? It's not clear how I would find $\sup_i\alpha_{ij}$ for fixed arbitrary $j$ first. Thanks.

share|cite|improve this question
If you read $\sup$ as "least upper bound", does that help? – cardinal Jan 16 '12 at 2:17
@cardinal I am thinking of $\sup$ as least upper bound, but I don't know how to conceptually find the least upper bound $\sup_i\ \alpha_{ij}$ if $j$ is some arbitrary fixed index. How can you know you have an upper bound of $\alpha_{ij}$ when it's not clear what $j$ is yet? – blackmoore Jan 16 '12 at 2:21
up vote 4 down vote accepted

Yes, it means exactly what you wrote.

The reason that

$$\sup_i\,\sup_j\ \alpha_{ij}=\sup_j\,\sup_i\ \alpha_{ij}$$

is that they're both equal to


For assume that

$$\sup_i\,\sup_j\ \alpha_{ij}\lt\sup_{i,j}\,\alpha_{ij}\;.$$

Then $\sup_i\,\sup_j\ \alpha_{ij}$ is not an upper bound for the $\alpha_{ij}$ (since there is no upper bound less than the supremum). Thus there is some $\alpha_{kl}$ greater than $\sup_i\,\sup_j\ \alpha_{ij}$. But this $\alpha_{kl}$ would make $\sup_j\alpha_{kj}$ be at least $\alpha_{kl}$, and thus $\sup_i\,\sup_j\alpha_{ij}$ would also be at least $\alpha_{kl}$, a contradiction.

Similarly, if

$$\sup_i\,\sup_j\ \alpha_{ij}\gt\sup_{i,j}\,\alpha_{ij}\;,$$

then some $\alpha_{kl}$ would have to be greater than $\sup_{i,j}a_{ij}$, which is impossible.

Note that this only works because both operations are suprema. If you take, say, the infimum with respect to $i$ and the supremum with respect to $j$, then it does matter in which order you perform those operations.

share|cite|improve this answer
Thanks joriki, this is clear to me. – blackmoore Jan 16 '12 at 2:42
You're welcome. – joriki Jan 16 '12 at 2:44

Both are the same as $$\sup_{(i,j)\in\mathbb{N}\times\mathbb{N}} \alpha_{i,j}.$$

share|cite|improve this answer
Somehow, I don't think that's going to be much help to the OP, unfortunately. – cardinal Jan 16 '12 at 2:19
Do you mind explaining more in detail why that is? – blackmoore Jan 16 '12 at 2:22

Suppose that $\lambda < \sup_i \sup_j \alpha_{i,j}$. Then there must be some $i$ so $\lambda < \sup_j \alpha_{i,j}$. Hence there is some $j$ so that $\lambda <\alpha_{i,j} \le \sup_{i,j} \alpha_{i,j}.$
We have $$\sup_i\, \sup_j \alpha_{i,j}\le \sup_{i,j}\alpha_{i,j}.$$

Now suppose that $\lambda < \sup_{i,j} \alpha_{i,j}$. Then there is some $(i,j)$ so $\lambda < \alpha_{i,j}\le \sup_i \sup_j \alpha_{i,j}.$ Since $\lambda$ was chosen arbitrarily, the reverse inequality holds.

share|cite|improve this answer
Did you mean to edit your previous answer? – cardinal Jan 16 '12 at 2:37

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.