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Let a triangle be defined as a three-element tuple $(a, b, c)$ where $a$, $b$, $c$ are positive real numbers and subjected to triangle inequality. Let $T = \{(a, b, c) | a, b, c \in \mathbb{R}_{>0}, a + b > c\}$. Prove that $T$ is uncountably infinite.

My proof scratch is the following:

Let $B = \{(a, b, c)| a, b, c \in \mathbb{R}_{>0}\}$. Notice that $B = \mathbb{R}_{>0} \times \mathbb{R}_{>0} \times \mathbb{R}_{>0} $. Since the Cartesian product of uncountable sets is uncountable, $B$ is uncountable.

The subset of an uncountable set is uncountable. By definition, $T$ is a subset of $B$.

Therefore $T$ is uncountably infinite. $$\tag*{$\blacksquare$}$$

UPDATE:

Based on the feedback from Anurag A, José Carlos Santos, and Patrick Stevens, the following is another attempt at the proof:

Theorem:

$T$ is uncountable.

Proof:

For an arbitrary but fixed triangle $(a, b, c)$ and an arbitrary positive real number $d$, $(da, db, dc)$ is a triangle.

Since there are uncountably infinite positive real number, the set of all $(da, db, dc)$ are uncountable.

Since the set of all $(da, db, dc)$ is uncountable and it is a subset of all triangles, the set of all triangles is uncountable. $$\tag*{$\blacksquare$}$$

UPDATE:

Based on the feedback from Patrick Stevens, the proof is simplified using injection.

Theorem:

$T$ is uncountable.

Proof:

Let $f(x) = (ax, bx, cx)$. For a fixed $u, v \in \mathbb{R}_{>0}$ and $(a, b, c) \in T$, suppose $f(u) = f(v)$, then:

$$\begin{equation}\begin{aligned} (au, bu, cu) &= (av, bv, cv)\\ \end{aligned}\end{equation}\tag{1}\label{eq1}$$

If we abuse the notation and treat both sides of $\eqref{eq1}$ as vectors: $$\begin{equation}\begin{aligned} \begin{bmatrix} au\\ bu\\ cu\\ \end{bmatrix} &= \begin{bmatrix} av\\ bv\\ cv\\ \end{bmatrix} \\ \begin{bmatrix} \frac{1}{a} \frac{1}{b} \frac{1}{c} \end{bmatrix} \begin{bmatrix} au\\ bu\\ cu\\ \end{bmatrix} &= \begin{bmatrix} \frac{1}{a} \frac{1}{b} \frac{1}{c} \end{bmatrix} \begin{bmatrix} av\\ bv\\ cv\\ \end{bmatrix}\\ 3u &= 3v\\ u &= v \end{aligned}\end{equation}\tag{2}\label{eq2}$$

By definition, $f$ is an injection from $\mathbb{R}_{>0}$ to $T$.

Therefore, $T$ has at least as many elements as $\mathbb{R}_{>0}$.

Since $\mathbb{R}_{>0}$ is uncountable, $T$ is uncountable. $$\tag*{$\blacksquare$}$$

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    $\begingroup$ $\Bbb{Z}$ is a subset of uncountable set $\Bbb{R}$. But it is countable. So just because $T$ is a subset of an uncountable set $B$ it need not be uncountable. $\endgroup$ – Anurag A Oct 27 '18 at 6:31
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Your proof is wrong, since, in general a subset of an uncountable set is not uncountable.

Not that if $(a,b,c)$ is a triangle, then any $(a+d,b+d,c+d)$ is also a triangle. This is enough to prove that the set of all triangles is uncountable.

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    $\begingroup$ Possibly even simpler: multiplication by a constant yields a triangle. (This is trivially obvious geometrically: zooming in on a triangle doesn't stop it being a triangle.) $\endgroup$ – Patrick Stevens Oct 27 '18 at 7:42
  • $\begingroup$ Yes, that's even simpler. $\endgroup$ – José Carlos Santos Oct 27 '18 at 7:49
  • $\begingroup$ @PatrickStevens I have rewrite the proof. Would you mind to check the proposition about $(da, db, dc)$ being uncountable? $\endgroup$ – Mys_721tx Oct 28 '18 at 7:54
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    $\begingroup$ @Mys_721tx You need to justify that $\{(da, db, dc) \mid d \in \mathbb{R}^{>0}\}$ is uncountable - you need to show that none of them are equal. Otherwise it's fine. $\endgroup$ – Patrick Stevens Oct 28 '18 at 7:57
  • $\begingroup$ @PatrickStevens I started working on that lemma and noticed that the entire proof can be simplified by showing that there is an injection from $\mathbb{R}_{>0}$ and $T$. $\endgroup$ – Mys_721tx Oct 28 '18 at 22:11

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