Transitivity of cardinality of sets

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 $f:\mathbb R \to (0,\infty)$ defined by $f(x)=e^x$

This is a bijection so $|\mathbb R | =|(0,\infty)|$ but $|(0,1) | =|(0,\infty)|$ then $|\mathbb R | =|(0,1)|$


i get the final conclusion but i'm not understanding where the (0,1) comes from.

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'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 $g : (0,\infty)\to (0,1)$ defined by $g(x)=\dfrac{1}{1+x}$

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 $\dfrac{1}{1-x}$ which ii think probabyl helps, thanks a bunch

Wtih that domain the codomain for that function is $(1,\infty)$, for a bijection.

sorry my mistake i misread, the function was $\dfrac{x}{1-x}$ my bad