Re: Topological sum

On May 20, 5:08 am, David Hartley <m...@xxxxxxxxxxx> wrote:
In message <1179632618.227598.121...@xxxxxxxxxxxxxxxxxxxxxxxxxxxx>,
fjbl...@xxxxxxxxx writes

On May 17, 7:08 am, bockermann <fstr...@xxxxxx> wrote:
Dear mathematicians,

are theretopologicalspaces A and B, such that the disjoint union
(coproduct)
A U A is homeomorphic to B U B, but A is not homeomorphic to B?

[I thought I posted this yesterday but it has not appeared. Sorry if
it's a duplicate.]

The answer is no.

First consider the case where A and B are discrete, so this is a
statement about sets and bijections. If A or B is finite, it's
obvious. If A and B infinite, it's a theorem of cardinal arithmetic
(using Zorn's lemma) that A U A is in bijective correspondence with A,
so the statement is true there as well.

Next, suppose A is a disjoint union of homeomorphic components, i.e. A
= X x D, X connected, D discrete. Show that B U B is a disjoint union
of components homeomorphic to X. A homeomorphism from A U A to B U B
gives a bijection on their respective components, and by the previous
case there is a bijection between the components of A and B, giving a
homeomorphism since those components are homeomorphic.

Finally, for general A, classify the components by homeomorphism
type. The existence of a homeomorphism will show that the components
of B U B admit a similar classification, and then the previous part
can be applied to each class.

In general, a topological space is not the disjoint union of its
components. E.g. the rationals with the usual topology.

Whoops. Thanks for setting me straight, and for the reference.

.

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