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:

$(1): \quad X$ contains an uncountable subset

$(2): \quad X$ is uncountable

$(3): \quad $ Every sequence of distinct points $(x_n)_{n \in \N}$ in $X$ omits at least one $x \in X$

$(4): \quad $ There is no surjection $\N \twoheadrightarrow X$

$(5): \quad X$ is infinite and there is no bijection $X \leftrightarrow \N$

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

$(6): \quad $ 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 $\hat f: \N \to Y$ defined as:

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

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

$\Box$

$(2) \implies (3)$

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.

$\Box$

$(3) \implies (4)$

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.

$\Box$

$(4) \implies (5)$

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

$\Box$

$(5) \implies (1)$

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 $S_1 = \operatorname {Im} \left({f}\right)$ 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 $S_{n} \subseteq \operatorname {Im} \left({f}\right)$, 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 $\operatorname {Im} \left({f}\right)$ 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.

$\Box$

$(6) \implies (1)$

$\blacksquare$