darkprince14
  • darkprince14
Prove: If A is an uncountable Set and B is equinumerous to A, then B is uncountable.
Mathematics
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SOLVED
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chestercat
  • chestercat
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darkprince14
  • darkprince14
@.Sam. Please Help..
anonymous
  • anonymous
Hmm, what is equinumerous?
darkprince14
  • darkprince14
if there exists a bijective function between the two..

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zzr0ck3r
  • zzr0ck3r
there is a 1-1 between them
anonymous
  • anonymous
Okay, I think proof by contradiction would work best.
darkprince14
  • darkprince14
I proved that B is infinite but stopped at proving on how are they not equinumerous to N...
darkprince14
  • darkprince14
I mean how B is not equinumerous to N, since A is already not equinumerous to N.
anonymous
  • anonymous
The negation of the statement is: "B is countable and A is uncountable and B is equinumerous to A" This is a contradiction.
darkprince14
  • darkprince14
wont the statement "B equinumerous to A" be also negated?
anonymous
  • anonymous
"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."
anonymous
  • anonymous
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.
zzr0ck3r
  • zzr0ck3r
\[\text{assume A is countable, and there exists a bijection from A to B } f:A\rightarrow B\\\text{assume by contradiction that B is countable, then there exists a bijection}\\\text{from B to N , }g:B\rightarrow N\\then\\g(f(a)):A\rightarrow N\text{ is a bijection} \\\text{thus A is countable, and this is a contradiction}\]
darkprince14
  • darkprince14
ooh, now I get it, thanks!!! :D
zzr0ck3r
  • zzr0ck3r
this rests on the fact that the composition of bijections, is a bijection
zzr0ck3r
  • zzr0ck3r
this is very easy to prove if you need it...

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