Difference between revisions of "Disjoint union topology"

Revision as of 18:33, 24 September 2016

Definition

Suppose [ilmath]\big((X_\alpha,\mathcal{J}_\alpha)\big)_{\alpha\in I} [/ilmath] be an indexed family of topological spaces that are non-empty^[1], the disjoint union topology is a topological space:

• with underlying set [ilmath]\coprod_{\alpha\in I}X_\alpha[/ilmath], this is the disjoint union of sets, recall [ilmath](x,\beta)\in\coprod_{\alpha\in I}X_\alpha\iff \beta\in I\wedge x\in X_\beta[/ilmath] and
• The topology where [ilmath]U\in\mathcal{P}(\coprod_{\alpha\in I}X_\alpha)[/ilmath] is considered open if and only if [ilmath]\forall \alpha\in I[X_\alpha\cap U\in\mathcal{J}_\alpha][/ilmath]^{[Note 1]} - be sure to notice the abuse of notation going on here.

TODO: Flesh out notes, mention subspace [ilmath]X_\alpha\times\{\alpha\} [/ilmath] and such

Claim 1: this is indeed a topology

Proof of claims

Notes

1. ↑ There's a very nasty abuse of notation going on here. First, note a set [ilmath]U[/ilmath] is going to be a bunch of points of the form [ilmath](x,\gamma)[/ilmath] for various [ilmath]x[/ilmath]s and [ilmath]\gamma[/ilmath]s ([ilmath]\in I[/ilmath]). There is no "canonical projection" FROM the product to the spaces, as this would not be a function!

References

1. ↑ Introduction to Topological Manifolds - John M. Lee

TODO: Investigate the need to be non-empty, I suspect it's because the union "collapses" in this case, and the space wouldn't be a part of union