4
$\begingroup$

I'm looking at a question out of Lee's Smooth Manifolds:

Show that a disjoint union of uncountably many copies of $\Bbb{R}$ is locally Euclidian and Hausdorff but not second countable.

My question is not about the problem itself, but rather what a common notation would be to denote this space. It just felt awkward going to write a solution to the problem, and writing the space as $M=\sqcup_{i\in I}M_i$ with $M_i\cong\Bbb{R}$. On the other hand, I could have gone with $\sqcup_{i\in I}\Bbb{R}$, but if I choose points $x,y\in M$ that doesn't give me a way to effectively describe which copies of $\Bbb{R}$ $x$ and $y$ lie in, respectively.

There's a possibility this question will get closed as off-topic, but I figured I'd try anyways. I know in the end it's down to personal preference, but surely there's some sort of standard.

$\endgroup$
4
$\begingroup$

I don't think there's any notation that is more standard than $\coprod_{i\in I}\mathbb{R}$ that makes it clear you are talking about a disjoint union of copies of $\mathbb{R}$. But if you want a nice way to be able to refer to the individual copies, you could just write it as $I\times\mathbb{R}$, with the $i$th copy being $\{i\}\times\mathbb{R}$. The disjoint union topology even coincides with the product topology if you give $I$ the discrete topology.

$\endgroup$
2
$\begingroup$

You understand that the usual implementation of disjoint union is as tagged elements (ordered pairs): $$ \bigsqcup_{i \in I} A_i = \bigcup_{i \in I} \{(i,x) \mid x \in A_i\} $$

So your $x$ and $y$ are pairs indicating, respectively, the index of their member of the disjoint union and element from that member.

Note that it is very difficult to distinguish this set from the set $I \times \Bbb{R}$ if all the $A_i$ are copies of $\Bbb{R}$. (It's just that the disjoint union inherits the union of the member topologies, not the product topology and using the Cartesian product notation could easily conflate the two in the reader's mind.)

$\endgroup$
2
$\begingroup$

You can be less formal by saying something like this, and everyone will understand you:

  • Let $\cup_{i \in I} R_i$ denote a disjoint union of copies $R_i$ of $\mathbb{R}$...
$\endgroup$

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.