Download MATH 4450 HOMEWORK SET 1, SOLUTIONS Problem 1 (2.8

MATH 4450 HOMEWORK SET 1, SOLUTIONS
Problem 1 (2.8): Suppose X and Y are sets, each of which has at least two elements. Show that X × Y
contains a subset that is not of the form A × B for any A ⊆ X, B ⊆ Y .
Let X have distinct elements x1 , x2 , and let Y have distinct elements y1 , y2 . (There may
be other elements in X and Y as well.) Note that, for any A ⊆ X and B ⊆ Y , A × B has
to be “rectangular”: if ha1 , b1 i and ha2 , b2 i are in A × B, then so are ha1 , b2 i and ha2 , b1 i.
With this in mind, suppose R = {hx1 , y1 i, hx2 , y2 i}. If R were some A × B, then hx1 , y2 i
would have to be in R. But then we’d have hx1 , y2 i = hx1 , y1 i or hx1 , y2 i = hx2 , y2 i. In the
first instance—by the ordered pair property—we have y1 = y2 ; in the second, x1 = x2 . But
we have assumed x1 6= x2 and y1 6= y2 , so this contradiction tells us R can’t be “rectangular.”
Problem 2 (3.6): Let f : X → Y , with A ⊆ X and B ⊆ Y . Show that f [A ∩ f −1 [B]] = f [A] ∩ B.
An element of the left-hand side (lhs) is an element y ∈ Y such that y = f (x) for some
x ∈ A ∩ f −1 [B]. Then y ∈ f [A] and y ∈ B; hence y is in the right-hand side (rhs).
For the reverse inclusion, suppose y is in the rhs. There is some x ∈ A such that y = f (x);
also, since y ∈ B, this x must also be in f −1 [B]. Thus x ∈ A ∩ f −1 [B], so y is in the lhs.
Problem 3 (4.3): For x, y ∈ R, define xRy by the condition that |x − y| < 1. Which of the relation properties
listed above (i.e., reflexivity, irreflexivity, symmetry, anti-symmetry, transitivity) pertain to R?
Since |x − x| = 0 < 1 and |x − y| = |y − x| always, we know R is reflexive and symmetric.
Reflexivity and irreflexivity are contradictory, so R is not irreflexive. The only relation that
is both symmetric and anti-symmetric is the identity relation (Why?), so anti-symmetry fails
for R. As for transitivity, we have 0R 12 and 21 R1, but it is not the case that 0R1. Thus R is
not transitive.
Problem 4 (5.2): Show that the tangent function f (x) = tan x defines a bijection between R and the interval
(−π/2, π/2).
Using basic ideas from Calculus, we have f ′ (x) = sec2 x, which is positive for −π/2 < x <
π/2. Hence f is one-one onto its image. Every real number is the tangent of a unique number
(angle) in (−π/2, π/2), so f is onto R. (Indeed, its inverse is the arctangent function.)
Problem 5 (5.3): Using the result of Problem 4, show that if a < b are real numbers then the interval (a, b)
is equinumerous with the real line.
Given a < b, the degree 1 function g(x) = mx+c, where m = π1 (b−a) and c = 21 (b+a), gives
a bijection from (−π/2, π/2) to (a, b). Since f (x) = tan x gives a bijection from (−π/2, π/2)
to R, the composition f ◦ g −1 : (a, b) → R is a bijection that witnesses the equinumerosity
of (a, b) and R.
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