Corollary 7.4.5

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


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

\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.


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


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.


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