# Switching order of supremum for doubly indexed sequence?

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.

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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

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

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

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.

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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 $$\sup_{(i,j)\in\mathbb{N}\times\mathbb{N}} \alpha_{i,j}.$$

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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.

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Did you mean to edit your previous answer? – cardinal Jan 16 '12 at 2:37