Show that (0, 1 ) is equivalent to [ 0, I ] and to IR and Prove that (0, 1 ) can

Question

Show that (0, 1 ) is equivalent to [ 0, I ] and to IR and Prove that (0, 1 ) can be put into one-to-one correspondence with the set of all functions f : N ---> {0, 1 } Show that any collection of pairwise disjoint, nonempty open intervals in 1R is at most countable however, If A is uncountable and B is countable. show that A and A B are equivalent. In particular, conclude that A B is uncountable and Show that the set of all real numbers in the interval (0, l ) whose base l0 decimal
expansion contains no 3s or 7 s is uncountable

Explanation / Answer

In every such open (non-trivial) interval there's at least one rational, and the union of all these rationals over that family is at most Q , and the cardinality of Q.

Or another (kind of) approach: as before with the rationals: if the family were uncountable, then taking the union of it would yield uncountable rationals

Once you have set up a map f:UQf:UQ such that f(I)Qf(I)Q and
f(I)If(I)I for each IUIU (and I think you might need the Axiom of Choice to do this; but I'm no expert on set theory so if someone more knowledgeable wants to chime in, I for one would welcome it . . .),
then injectivity follows from the fact that members of UU are disjoint, for if I1,I2UI1,I2U
with f(I1)=f(I2)=qQf(I1)=f(I2)=qQ,
we have qI1qI1 and qI2qI2, contradicting I1I2=I1I2=.
Since Image(f)QImage(f)Q, it is countable by virtue of being a subset of a countable set. It seems to me that that the OP's argument for injectivity is basically the same as mine, but I don't quite see where one needs the fact that a (countable) union of countable sets is countable.

A is uncountable and B is countable. show that A and A B are equivalent:

Suppose that AB is at most countable. Then, A = (AB)B is a union of two at most countable sets, hence A is at most countable, contrary to our hypothesis.

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