Closed Interval in Reals is Uncountable

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Let $a, b$ be extended real numbers such that $a < b$.

Then the closed interval $\set {x \in \R : a \le x \le b} \subseteq \R$ is uncountable.


First suppose that $a, b \in \R$.

We have that the unit interval is uncountable.

Let $f: \closedint 0 1 \to \closedint a b$ such that $\map f x = a + \paren {b - a} x$.

Then if $\map f {x_1} = \map f {x_2}$, we have:

$a + \paren {b - a} x_1 = a + \paren {b - a} x_2$

Because $b > a$:

$b - a > 0$

so this implies:

$x_1 = x_2$

Therefore $f$ is injective.

Now if $\closedint a b$ were countable, there would be an injective function $g : \closedint a b \to \N$.

By Composite of Injections is Injection, this implies that $g \circ f$ is an injection, so $\closedint 0 1$ is countable, a contradiction.

If one or both of $a, b$ is (are) not real, then we can pick a closed interval $\closedint c d \subseteq \closedint a b$ with $c, d \in \R, c < d$.

We know by the above that $\closedint c d$ is uncountable.

Let $S = \set {x \in \R : a \le x \le b}$.

The identity function $I_\R : \closedint c d \to S$ is trivially injective.

If $S$ were uncountable, we would have an injective function $g : S \to \N$.

Then by Composite of Injections is Injection we would have an injection $\closedint c d \to \N$ given by $g \circ I_\R$.

This was shown above to be false, so $S$ is uncountable.