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# Transitivity of cardinality of sets watch

1. Suppose |A|=|B| and |B|=|C| then there are bijections f:A->B and g:B->C and since gof:A->C inherits bijectivity from f and g then |A|=|C|.

So for some example my teacher gave he said

consider defined by

This is a bijection so but then

i get the final conclusion but i'm not understanding where the (0,1) comes from.
2. Did you have a previous example of a bijection mapping (0, 1) to (0, infty)? (It is not at all hard to find one),
3. (Original post by will'o'wisp2)
Suppose |A|=|B| and |B|=|C| then there are bijections f:A->B and g:B->C and since gof:A->C inherits bijectivity from f and g then |A|=|C|.

So for some example my teacher gave he said

consider defined by

This is a bijection so but then

i get the final conclusion but i'm not understanding where the (0,1) comes from.
There's nothing in your post to tell us. Perhaps it's something that was discussed in class/lectures.

We could create a relationshp.

E.g. Consider defined by

This is a bijection, and a "g" we can use in the transitivity.
4. (Original post by DFranklin)
Did you have a previous example of a bijection mapping (0, 1) to (0, infty)? (It is not at all hard to find one),
There is a function on the previous sheet of paper which was the function
5. (Original post by ghostwalker)
There's nothing in your post to tell us. Perhaps it's something that was discussed in class/lectures.

We could create a relationshp.

E.g. Consider defined by

This is a bijection, and a "g" we can use in the transitivity.
oh right i see, yes there was a bijective mapping from (0,1)->(0,infinity) for a function which ii think probabyl helps, thanks a bunch
6. (Original post by will'o'wisp2)
oh right i see, yes there was a bijective mapping from (0,1)->(0,infinity) for a function which ii think probabyl helps, thanks a bunch
Wtih that domain the codomain for that function is , for a bijection.
7. (Original post by ghostwalker)
Wtih that domain the codomain for that function is , for a bijection.
8. (Original post by will'o'wisp2)
Yep, that will do it.

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