Re: Question on union of open sets

On 02-03-2008 9:47, William Elliot wrote:

On 1 Mar, 23:18, José Carlos Santos <jcsan...@xxxxxxxx> wrote:
Let A be your set. For each _x_ in A, let I_x be the union of all open
intervals contained in A of which _x_ is an element. Consider the set

{ I_x : x in A }

Each element of this set is open, any two elements of this set are
disjoint and their union is A.
Right! If for another y in A the intersection I_x and I_y is non
empty, we have I_x subset of I_x U I_y thus, (because I_x is
"maximal") I_x = I_y.

Not so fast. You have to show I_x exists.

No he doesn't. The set I_x always exists and it is empty if and only if
_x_ is not an interior point of A.

That for all x in A, there is
an open interval U such that x in U subset A. This is not difficult, but
is an essential step. Otherwise, the same prove could be used for closed
sets and other sets, neither closed nor open.

In general, the proof that I described proves that the interior of A can
be written as an union of a disjoint family of open intervals.

Best regards,

Jose Carlos Santos
.

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