Question

Prove: If A is an uncountable Set and B is equinumerous to A, then B is uncountable.

Answer 1

@.Sam.
Please Help..

Answer 2

Hmm, what is equinumerous?

Answer 3

if there exists a bijective function between the two..

Answer 4

there is a 1-1 between them

Answer 5

Okay, I think proof by contradiction would work best.

Answer 6

I proved that B is infinite but stopped at proving on how are they not equinumerous to N...

Answer 7

I mean how B is not equinumerous to N, since A is already not equinumerous to N.

Answer 8

The negation of the statement is:
"B is countable and A is uncountable and B is equinumerous to A"

This is a contradiction.

Answer 9

wont the statement "B equinumerous to A" be also negated?

Answer 10

"If A is an uncountable Set and B is equinumerous to A, then B is uncountable."\[
p\implies q \iff \neg \,p\vee q
\]" NOT(A is an uncountable Set and B is equinumerous to A) OR B is uncountable."
\[
\neg(\neg \,p\vee q) \iff p\wedge \neg \,q
\]So negating gives:
"A is an uncountable Set and B is equinumerous to A AND NOT (B is uncountable.)"
Or simply:
"A is an uncountable Set and B is equinumerous to A and B is countable."

Answer 11

The proof is simple. We can map to B, so we can map through B to A. But if A is uncountable this shouldn't be possible.

Answer 12

\[\text{assume A is countable, and there exists a bijection from A to B } f:A\rightarrow B\\\text{assume by contradiction that B is countable, then there exists a bijection}\\\text{from B to N , }g:B\rightarrow N\\then\\g(f(a)):A\rightarrow N\text{ is a bijection} \\\text{thus A is countable, and this is a contradiction}\]

Answer 13

ooh, now I get it, thanks!!! :D

Answer 14

this rests on the fact that the composition of bijections, is a bijection

Answer 15

this is very easy to prove if you need it...