3. (a) Show that the intervals (0, 1) and (4, 9) are equinumerous by finding a b

Question

3.

(a) Show that the intervals (0, 1) and (4, 9) are equinumerous by finding a bijection f:(0, 1) -> (4, 9).

(b) Suppose s < t.

Prove that the intervals (0, 1) and (s, t) are equinumerous by finding a bijection f:(0, 1) -> (s, t).

(c) Suppose a < b and c < d. By part (b), there exist bijections g:(0, 1) -> (a, b) and h:(0, 1) -> (c, d). Show that open intervals (a, b) and (c, d) are equinumerous by finding a bijection f: (a, b) -> (c, d), where f is an appropriate composition of functions involving functions g, h, and/or their inverses.

Explanation / Answer

The bijection is (f(x) = 4 + 5x), such that (f:(0,1) ightarrow (4,9)).

We can generalize the answer for part (a) as followed: (f(x) = s + (t - s)x), where (f:(0,1) ightarrow (s,t)).

We have that (g(x) = a + (b - a)x) and (h(x) = c + (d - c)x), such that (g:(0,1) ightarrow (a,b)) and (h:(0,1) ightarrow (c,d)) respectively. Since (g) is a bijection, it has an inverse (g^{-1}), which is

[g^{-1}(x) = dfrac{1}{b - a}(x - a)]

Therefore, we write

[f(x) = (hcirc g^{-1})(x) = h(g^{-1}(x))]

which is

[f(x) = c + (d - c)g^{-1}(x) = c + dfrac{d - c}{b - a}(x - a)]

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