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In the paper An Unusual Proof that the Reals are Uncountable a proof of the uncountability of the reals is given which is adapted from one of Bourbaki's proofs in "Fonctions d'une variable réelle". Let me give it here:

Suppose $R$ was countable. Then there is a function $a(x) : R → R$ such that:

1. $a(x) > 0$ for all $x$,

2. the sum of the $a(x)$ on any finite set is $≤ 1.$

(take $a(x) = 2^{-n}$ if $x$ is the $n$’th element in the counting).

Now, define for any $S\subseteq R$,

$m(S)=$ the supremum of the sums of $a(x)$ on finite subsets of $S$.

Then surely $0 ≤ m(S) ≤ 1$ for any $S$. Define:

$c := \sup\{x: m(-\infty, x)>x \}$

Since $a(c) > 0,$ there is a $y$ such that $y > c − a(c)$ and $m( − ∞, y)> y$, thus $y ≤ c$. Now, since $y ≤ c$, $( − ∞, y)$ does not contain $c$. But $( −∞, y + a(c))$ contains $c$ because $y > c − a(c).$ So by the definition of $m(S)$,

$m( − ∞, y + a(c)) ≥ m( − ∞, y) +a(c) > y + a(c)$,

But $y + a(c) > c$ a contradiction.

In the paper An Unusual Proof that the Reals are Uncountable a proof of the uncountability of the reals is given which is adapted from one of Bourbaki's proofs in "Fonctions d'une variable réelle". Let me give it here:

Suppose $R$ was countable. Then there is a function $a(x) : R → R$ such that:

1. $a(x) > 0$ for all $x$,

2. the sum of the $a(x)$ on any finite set is $≤ 1.$

(take $a(x) = 2^{-n}$ if $x$ is the $n$’th element in the counting).

Now, define for any $S\subseteq R$,

$m(S)=$ the supremum of the sums of $a(x)$ on finite subsets of $S$.

Then surely $0 ≤ m(S) ≤ 1$ for any $S$. Define:

$c := \sup\{x: m(-\infty, x)>x \}$

Since $a(c) > 0,$ there is a $y$ such that $y > c − a(c)$ and $m( − ∞, y)> y$, thus $y ≤ c$. Now, since $y ≤ c$, $( − ∞, y)$ does not contain $c$. But $( −∞, y + a(c))$ contains $c$ because $y > c − a(c).$ So by the definition of $m(S)$,

$m( − ∞, y + a(c)) ≥ m( − ∞, y) +a(c) > y + a(c)$,

But $y + a(c) > c$ a contradiction.

In the paper An Unusual Proof that the Reals are Uncountable a proof of the uncountability of the reals is given which is adapted from one of Bourbaki's proofs in "Fonctions d'une variable réelle". Let me give it here:

Suppose $R$ was countable. Then there is a function $a(x) : R → R$ such that:

1. $a(x) > 0$ for all $x$,

2. the sum of the $a(x)$ on any finite set is $≤ 1.$

(take $a(x) = 2^{-n}$ if $x$ is the $n$’th element in the counting).

Now, define for any $S\subseteq R$,

$m(S)=$ the supremum of the sums of $a(x)$ on finite subsets of $S$.

Then surely $0 ≤ m(S) ≤ 1$ for any $S$. Define:

$c := \sup\{x: m(-\infty, x)>x \}$

Since $a(c) > 0,$ there is a $y$ such that $y > c − a(c)$ and $m( − ∞, y)> y$, thus $y ≤ c$. Now, since $y ≤ c$, $( − ∞, y)$ does not contain $c$. But $( −∞, y + a(c))$ contains $c$ because $y > c − a(c).$ So by the definition of $m(S)$,

$m( − ∞, y + a(c)) ≥ m( − ∞, y) +a(c) > y + a(c)$,

But $y + a(c) > c$ a contradiction.

The following is another proof which uses forcing (see also Forcing as a replacement of induction and diagonal arguments):

Suppose that $\mathbb{R}$ is countable. Let $\mathbb{P}$ consists of pairs $(A, n),$ where $A$ is a finite subset of $\mathbb{R}$, and $n$ is a natural number, so that $0.n \notin A.$ Say $(B, m)$ extends $(A, n)$ iff $A \subseteq B$ and $m$ is a continuation of $n$ (for example 235 is a continuation of 2 or 23 or 235, while it is not a continuation of 3 or 24, ..). For each real $r$, the set $A_r=\{(A, n): r\in A \}$ is clearly dense. Let $G$ be a filter meeting all $A_r$'s (recall that by our assumption there are only countably many reals).

Now let $t=0.a_1a_2...,$ where $a_1a_2...$ is a continuation of all $n$'s appearing in some condition in $G$. Clearly $t\neq r$ for any real $r$, a contradiction.

In the paper An Unusual Proof that the Reals are Uncountable a proof of the uncountability of the reals is given which is adapted from one of Bourbaki's proofs in "Fonctions d'une variable réelle". Let me give it here:

Suppose $R$ was countable. Then there is a function $a(x) : R → R$ such that:

1. $a(x) > 0$ for all $x$,

2. the sum of the $a(x)$ on any finite set is $≤ 1.$

(take $a(x) = 2^{-n}$ if $x$ is the $n$’th element in the counting).

Now, define for any $S\subseteq R$,

$m(S)=$ the supremum of the sums of $a(x)$ on finite subsets of $S$.

Then surely $0 ≤ m(S) ≤ 1$ for any $S$. Define:

$c := \sup\{x: m(-\infty, x)>x \}$

Since $a(c) > 0,$ there is a $y$ such that $y > c − a(c)$ and $m( − ∞, y)> y$, thus $y ≤ c$. Now, since $y ≤ c$, $( − ∞, y)$ does not contain $c$. But $( −∞, y + a(c))$ contains $c$ because $y > c − a(c).$ So by the definition of $m(S)$,

$m( − ∞, y + a(c)) ≥ m( − ∞, y) +a(c) > y + a(c)$,

But $y + a(c) > c$ a contradiction.

In the paper An Unusual Proof that the Reals are Uncountable a proof of the uncountability of the reals is given which is adapted from one of Bourbaki's proofs in "Fonctions d'une variable réelle". Let me give it here:

Suppose $R$ was countable. Then there is a function $a(x) : R → R$ such that:

1. $a(x) > 0$ for all $x$,

2. the sum of the $a(x)$ on any finite set is $≤ 1.$

(take $a(x) = 2^{-n}$ if $x$ is the $n$’th element in the counting).

Now, define for any $S\subseteq R$,

$m(S)=$ the supremum of the sums of $a(x)$ on finite subsets of $S$.

Then surely $0 ≤ m(S) ≤ 1$ for any $S$. Define:

$c := \sup\{x: m(-\infty, x)>x \}$

Since $a(c) > 0,$ there is a $y$ such that $y > c − a(c)$ and $m( − ∞, y)> y$, thus $y ≤ c$. Now, since $y ≤ c$, $( − ∞, y)$ does not contain $c$. But $( −∞, y + a(c))$ contains $c$ because $y > c − a(c).$ So by the definition of $m(S)$,

$m( − ∞, y + a(c)) ≥ m( − ∞, y) +a(c) > y + a(c)$,

But $y + a(c) > c$ a contradiction.

In the paper An Unusual Proof that the Reals are Uncountable a proof of the uncountability of the reals is given which is adapted from one of Bourbaki's proofs in "Fonctions d'une variable réelle". Let me give it here:

Suppose $R$ was countable. Then there is a function $a(x) : R → R$ such that:

1. $a(x) > 0$ for all $x$,

2. the sum of the $a(x)$ on any finite set is $≤ 1.$

(take $a(x) = 2^{-n}$ if $x$ is the $n$’th element in the counting).

Now, define for any $S\subseteq R$,

$m(S)=$ the supremum of the sums of $a(x)$ on finite subsets of $S$.

Then surely $0 ≤ m(S) ≤ 1$ for any $S$. Define:

$c := \sup\{x: m(-\infty, x)>x \}$

Since $a(c) > 0,$ there is a $y$ such that $y > c − a(c)$ and $m( − ∞, y)> y$, thus $y ≤ c$. Now, since $y ≤ c$, $( − ∞, y)$ does not contain $c$. But $( −∞, y + a(c))$ contains $c$ because $y > c − a(c).$ So by the definition of $m(S)$,

$m( − ∞, y + a(c)) ≥ m( − ∞, y) +a(c) > y + a(c)$,

But $y + a(c) > c$ a contradiction.

The following is another proof which uses forcing (see also Forcing as a replacement of induction and diagonal arguments):

Suppose that $\mathbb{R}$ is countable. Let $\mathbb{P}$ consists of pairs $(A, n),$ where $A$ is a finite subset of $\mathbb{R}$, and $n$ is a natural number, so that $0.n \notin A.$ Say $(B, m)$ extends $(A, n)$ iff $A \subseteq B$ and $m$ is a continuation of $n$ (for example 235 is a continuation of 2 or 23 or 235, while it is not a continuation of 3 or 24, ..). For each real $r$, the set $A_r=\{(A, n): r\in A \}$ is clearly dense. Let $G$ be a filter meeting all $A_r$'s (recall that by our assumption there are only countably many reals).

Now let $t=0.a_1a_2...,$ where $a_1a_2...$ is a continuation of all $n$'s appearing in some condition in $G$. Clearly $t\neq r$ for any real $r$, a contradiction.