Part b should be proven by showing that there is a bijection

Part (b) should be proven by showing that there is a bijection between [0,1] and [a,b]

Exercise 1.4.3. (a) Show that the interval [0, 1] is uncountable. (b) Show that the interval [a, b] is uncountable.

Solution

(a)

Cantor\'s diagonal argument is a mathematical proof that there are infinite sets which cannot be put into one-to-one correspondence with the infinite set of natural numbers. Such sets are now known as uncountable sets.

Proof

The bijection f, which we have assumed to exist, can map any positive integer to a value in [0,1] (and since it\'s a bijection, none of the points in [0,1] are left over). Also, the points in [0,1] can be treated as a number line so that each point on the line is a value between 00 and 11 which we can write out as a decimal.

Imagine writing these decimals out in order according to the integer they map to. So perhaps 10.5 and 22/3 and 31/:

1: 0.50000000...
2: 0.66666666...
3: 0.31830988...
..etc..
Now read down the diagonal (marked in bold above) and pick a different digit than what you see. For instance, if you see a 55, use \"66\", if you see anything else use \"55\". That gives us a series of digits... in this case it starts out \"0.655...0.655...\".

Clearly, this series of digits is not in the list, since it differs from each item in the list by at least one decimal place. Clearly it is a number in the range [0,1]. Since we used 55 and 66 it also doesn\'t have a repeating \"99999...99999...\" or \"00000...00000...\" (which is a way 22 different sequences of digits could represent the same number). So our assumption that f was a bijection must have been false.

Therefore [0,1] is uncountable.

(b)

[a, b] interval implies that it is set of all real numbers R. So need to prove that R is uncountable.

Proof

If R were countable, then [0, 1] would be a subset of a countable set and would be countable. Since [0, 1] is not countable, the result follows. Since any nonempty interval of real numbers can be put into 1-1 correspondence with [0, 1], every nonempty interval of real numbers is uncountable.

First need to show a 1-1 correspondence from [0, 1] to [0, b a], and then another one from [0, b a] to [a, b].

Let use conisder A= [0, 1], B= [0, b a]

If f is a 1-1 correspondence between A and B, then f associates every element of B with a unique element of A (at most one element of A because it is 1-1, and at least one element of A because it is onto). That is, for each element b B there is exactly one a A so that the ordered pair (a, b) f. Since f is a function, for every a A there is exactly one b B such that (a, b) f. Thus, f “pairs up” the elements of A and the elements of B. So when such a function f exists, we say A and B can be put into 1-1 correspondence. So 1-1 correspondence exists from [0, 1] to [0, b a]

Let use conisder A=[0, b a] to [a, b]

If f is a 1-1 correspondence between A and B, then f associates every element of B with a unique element of A (at most one element of A because it is 1-1, and at least one element of A because it is onto). That is, for each element b B there is exactly one a A so that the ordered pair (a, b) f. Since f is a function, for every a A there is exactly one b B such that (a, b) f. Thus, f “pairs up” the elements of A and the elements of B. So when such a function f exists, we say A and B can be put into 1-1 correspondence. So 1-1 correspondence exists from [0, b a] to [a, b]

Hence [a,b] is uncountable.