Difference between revisions of "Disjoint union topology"
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* The [[topology]] where {{M|U\in\mathcal{P}(\coprod_{\alpha\in I}X_\alpha)}} is considered [[open set|open]] ''if and only if'' {{M|1=\forall \alpha\in I[X_\alpha\cap U\in\mathcal{J}_\alpha]}}<ref group="Note">There's a very nasty abuse of notation going on here. First, note a set {{M|U}} is going to be a bunch of points of the form {{M|(x,\gamma)}} for various {{M|x}}s and {{M|\gamma}}s ({{M|\in I}}). There is no "canonical projection" FROM the product to the spaces, as this would not be a [[function]]!</ref> - be sure to notice the abuse of notation going on here. | * The [[topology]] where {{M|U\in\mathcal{P}(\coprod_{\alpha\in I}X_\alpha)}} is considered [[open set|open]] ''if and only if'' {{M|1=\forall \alpha\in I[X_\alpha\cap U\in\mathcal{J}_\alpha]}}<ref group="Note">There's a very nasty abuse of notation going on here. First, note a set {{M|U}} is going to be a bunch of points of the form {{M|(x,\gamma)}} for various {{M|x}}s and {{M|\gamma}}s ({{M|\in I}}). There is no "canonical projection" FROM the product to the spaces, as this would not be a [[function]]!</ref> - be sure to notice the abuse of notation going on here. | ||
{{Todo|Flesh out notes, mention subspace {{M|X_\alpha\times\{\alpha\} }} and such}} | {{Todo|Flesh out notes, mention subspace {{M|X_\alpha\times\{\alpha\} }} and such}} | ||
| − | + | '''Claim 1: ''' this is indeed a topology | |
| + | ==Proof of claims== | ||
| + | {{Requires proof|msg=Actually surprisingly easy to prove, done on paper. page 1, 7/8/2016, Intro to top manifolds notes. Filed}} | ||
==Notes== | ==Notes== | ||
<references group="Note"/> | <references group="Note"/> | ||
Revision as of 04:50, 7 August 2016
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Contents
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
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Actually surprisingly easy to prove, done on paper. page 1, 7/8/2016, Intro to top manifolds notes. Filed
Notes
- ↑ 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
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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
