Corollary 7.4.4

##### Claim

The set $\mathbb{R}$ of real numbers is uncountable.

**Proof**

For sake of contradiction, we will assume that $\mathbb{R}$ is countable. By the definition of countable there exists a 1-1 correspondence $f:\mathbb{N}\to\mathbb{R}$. In order to show our contradiction we must construct a new function:

$f*:X\to(1,0)$ via

$f*(n)=f(n)$

By assuming $f*$ to be onto and one to one we can see that $cardx=card(0,1)$. Therefore $(0,1)$ is countable, however, by theorem 7.4.3, $(0,1)$ is uncountable. So by contradiction $\mathbb{R}$ is uncountable.

**Proof**

Since $(0,1)$ is a subset of $\mathbb{R}$ and $(0,1)$ is uncountable by theorem 7.4.3, $\mathbb{R}$ must be uncountable.

$\blacksquare$

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page revision: 21, last edited: 10 Dec 2009 19:26

Use a proof by contradiction. Assume that $\mathbb{R}$ is countable. Then by definition, there exists a 1-1 correspondence $f: \mathbb{N}\to \mathbb{R}$. We'd like to use the fact that the interval $(0,1)$ is uncountable (see Theorem 7.4.3) to get a contradiction. Let $X=f^{-1}((0,1))$. Define a new function $f^*:X\to (0,1)$ via $f^*(n)=f(n)$ (this is similar to what we did in the proof of Theorem 5.1.16). Show that $f^*$ is onto. What does Theorem 7.3.5 say in this case?

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