Sufficient Conditions for Uncountability

Theorem
Let $X$ be a set.

Recall that $X$ is uncountable if there is no injection $X \hookrightarrow \N$.

The following are equivalent:


 * 1) $X$ contains an uncountable subset
 * 2) $X$ is uncountable
 * 3) Every sequence of distinct points $(x_n)_{n \in \N}$ in $X$ omits at least one $x \in X$
 * 4) There is no surjection $\N \twoheadrightarrow X$
 * 5) $X$ is infinite and there is no bijection $X \leftrightarrow \N$

Assuming the Continuum Hypothesis holds, we also have the equivalent uncountability condition:


 * 6. There exist extended real numbers $a < b$ and a surjection $X \to [a,b]$

Proof
1. $\implies$ 2.

Suppose that there is an injection $f:\N \hookrightarrow X$.

Let $Y \subseteq X$ be uncountable.

Then $\hat f: \N \to Y$,

$\hat f = \left\{(n,x) \in f : x \in Y \right\}$

is an injection $\N \hookrightarrow Y$, a contradiction.

2. $\implies$ 3.

Suppose that $(x_n)_{n \in \N}$ is a sequence such that every $x \in X$ equals $x_n$ for some $n$.

Since the $x_n$ are distinct, this $n$ is unique.

Let $f : X \to \N$ be defined by $f(x) = n$, where $n$ is the unique natural number such that $x_n = x$.

Then $f$ is injective because $n = m$ implies that $x_n = x_m$.

So there is an injection $X \hookrightarrow \N$, a contradiction.

3. $\implies$ 4.

Suppose that there is a surjection $f:\N \to X$.

Therefore every $x \in X$ is $f(n)$ for some $n \in \N$.

Define $g: X \to \N$ by $g(x) = \inf\{ n \in \N : f(n) = x \}$.

Then if $g(x_1) = g(x_2) = n$, we have $f(n) = x_1$ and $f(n) = x_2$.

So by the definition of a function, $x_1 = x_2$.

Therefore $g$ is an injection $X \to \N$, a contradiction.

4. $\implies$ 5.

Since there is no surjection $\N \to X$, and a bijection $\N \to X$ must be surjective, there is no such map.

5. $\implies$ 1.$\newcommand{\im}{\operatorname{im}}$

Suppose $f : X \hookrightarrow \N$ is an injection.

We essentially relabel the image of $f$ to obtain a bijection, and thus a contradiction.

Construct a map $g : X \to \N$ as follows. Let $S_1 = \im(f)$ and


 * $ x_1 = f^{-1} \left( \inf\{ n \in \N : n \in S_1 \} \right)$

and define $f(x) = 1$. Note that $x$ is a singleton because $f$ is injective.

Given $S_{n} \subseteq \im(f)$, let


 * $ x_n = f^{-1} \left( \inf\{ n \in \N : n \in S_n \} \right)$

and set $S_{n+1} = S_n \backslash x_n$.

The mapping


 * $n \stackrel{f^{-1}}{\longrightarrow} x_k \stackrel{g}{\longrightarrow} k$

defines a bijection from $\im(f)$ to $\N$, so $g$ is injective because $f$ is.

Furthermore $g$ assigns each $n \in \N$ to some $x \in X$ (because $X$ is infinite) so $g$ is surjective.

Thus $g$ is a bijection, a contradiction.

6. $\iff$ 1.