I am having a little trouble trying to prove that any nonempty open set in $\mathbb{R}$ is uncountable. Here is what I have so far:
Let $A$ be a nonempty subset of $\mathbb{R}$. Then for each $x \in A$, there exists an open interval $I=(a,b)$ such that $x \in I \subseteq A$.
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Since I is uncountable and $I \subseteq A$, it follows that $A$ is uncountable.
I am trying to fill in the … by showing that the open interval $(a,b)$ is uncountable. I am familiar with Cantor diagonalization and I've used it to prove that $(0,1)$ is uncountable, but I am unsure of how to set up a Cantor diagonal when I don't know what the boundaries of the interval are. I've read suggestions on other questions to set up a bijection from $(0,1)$ to $(a,b)$ but I'm not sure how to go about that either.
Best Answer
Instead of setting up Cantor diagonalization on $A$, find a surjective map $\phi$ from $A$ to $(0,1)$.
To do this, just pick $x \in A$. It is an interior point, so there is an interval $(a,b)$ around $x$ which is contained in $A$.
Now, $\phi$ is easy to construct : note that any $c \in (a,b)$ is of the form $ta + (1-t)b$ for some $t \in (0,1)$ (that is, $c$ is somewhere between $a$ and $b$, and $t$ is a measure of how far it is from each one), so let $\phi(c) = t$. Finally, for any $d \in A \setminus (a,b)$, let $\phi(d) = \frac 12$.
EDIT : An explicit description of $\phi$ is $\phi(c) = \frac{c-a}{b-a}$.
It is easy to see that $\phi$ is surjective : in fact, just $\phi$ restricted to $(a,b)$ is surjective.
Now if $A$ were countable, then so is $(a,b)$, because $(a,b)$ is a subset of $A$. Let $\psi$ be a bijection from the natural numbers to $(a,b)$. Then $\phi|_{(a,b)} \circ \psi$ is a bijective map from $\mathbb N$ to $(0,1)$.
This is a contradiction, since $(0,1)$ is uncountable, so there cannot be a bijective map from $\mathbb N$ to $(0,1)$.