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I'm confused. Two finite sets (call them A = [a, b] and C = [c, d]) are equivalent if there exists a 1-1 bijection from A to C. But the bijection exists iff A has the same number of elements as C.

So am I correct in saying that two finite sets can't be equivalent unless they have the same number of elements?

I've been asked this: Let a < b and c < d. Show [a, b] is equivalent to [c, d].

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    @chrislegend: No, $[0,10]$, as it happens, has "the same number" of elements as $[0,100]$; where "same number" means "there is$a$bijection between them". Infinity can be counterintuitive at first (one tends to develop a bit of intuition for it with practice)! That's the point of this exercise.2011-09-27

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We will construct the desired bijection geometrically, in steps.

(i) The function $f$ defined by $f(x)=x-a$ is a bijection from $[a,b]$ to $[0,b-a]$. This function $f$ is called a translation. (It translates, meaning shifts, everything over by an amount $-a$.)

(ii) We now map the interval $[0,b-a]$ bijectively to $[0, d-c]$. This can be done by scaling by the factor $\frac{d-c}{b-a}$ , that is, by applying the function $g$, where $g(x)=\frac{d-c}{b-a}x$.

(iii) After we have applied the transformation (i) and then (ii), the interval $[a,b]$ is mapped to the interval $[0,d-c]$. Now let $h(x)=x+c$. The translation $h$ maps $[0,d-c]$ bijectvely to $[c,d]$.

Finally, let $W(x)= h(g(f(x)))$. Then $W$ maps $[a,b]$ bijectively to $[c,d]$.

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    It is$a$clear and useful way to see things, a mixture of geometric and kinematic/kinesthetic. Beside pictures in the head, sometimes mathematics involves feeling things slotting into place, more muscular action than pattern.2011-09-28