Prove that any bounded open subset of R is the union of disjoint ope.pdf
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The document demonstrates that any bounded open subset of the real numbers can be expressed as a union of disjoint open intervals. It employs concepts from topology, including locally connected spaces and components, to establish that such subsets are composed of connected and open intervals. The conclusion is supported by the argument that there are at most countably many open intervals due to the presence of rational numbers within each interval.
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Prove that any bounded open subset of R is the union of disjoint ope.pdf
- 1. Prove that any bounded open subset of R is the union of disjoint open intervals. Please be thorough in your explantion. Thanks! Solution let O be an open subset of a locally connected space X. Let C be a component of O (as a (sub)space in its own right). Let x?C. Then let Ux be a connected neighbourhood of x in X such that Ux?O, which can be done as O is open and the connected neighbourhoods form a local base. Then Ux,C?O are both connected and intersect (in x) so their union Ux?C?O is a connected subset of O containing x, so by maximality of components Ux?C?C. But then Ux witnesses that x is an interior point of C, and this shows all points of Care interior points, hence C is open (in either X or O, that's equivalent). Now R is locally connected (open intervals form a local base of connected sets) and so every open set if a disjoint union of its components, which are open connected subsets of R, hence are open intervals (potentially of infinite "length", i.e. segments). That there are countably many of them at most, follows from the already given "rational in every interval" argument.