Corollary 7.4.5

Claim

Let $a, b \in \mathbb{R}$ with $a<b$. Then $(a,b), [a,b], (a,b]$, and $[a,b)$ are all uncountable.

Proof

The function $f : (0,1) \to (a,b)$ given by

(1)

\begin{equation} f(x)=a+x(b-a) \end{equation}

is 1-1 and onto. Because $(0,1)$ is uncountable, $(a,b)$ is also uncountable. The interval $(a,b)$ is a subset of $[a,b), (a,b],$ and $[a,b]$, so they are all uncountable. Also, $(0,1)$ is in $\mathbb {R}$, so $\mathbb {R}$ is uncountable. In conclusion, by exercise 7.1.2(2), if the $\text {card } (0,1)$ is uncountable then the $\text {card } (a,b)$ is uncountable.

$\blacksquare$

Lemma:
If $x$ is uncountable and $x \subseteq y$, then $y$ is uncountable.

proof

Assume $x$ is uncountable and $x \subseteq y$. For sake of a contradiction, assume that $y$ is countable. By Theorem 7.1.5, x is countable. So, by contradiction $y$ is uncountable.

$\blacksquare$

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Hints

Dana Ernst 03 Dec 2009 14:29

As in Corollary 7.4.4, use a proof by contradiction. In each case, assume the interval is countable. Appeal to Exercise 7.1.2(2) to conclude that each of these intervals has the same cardinality as an interval that you already know something about.