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..
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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.
