# Proof that the number of triangles is uncountably infinite

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. $$\tag*{\blacksquare}$$

• $\Bbb{Z}$ 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. – Anurag A Oct 27 '18 at 6:31

## 1 Answer

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.

• 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.) – Patrick Stevens Oct 27 '18 at 7:42
• Yes, that's even simpler. – José Carlos Santos Oct 27 '18 at 7:49
• @PatrickStevens I have rewrite the proof. Would you mind to check the proposition about $(da, db, dc)$ being uncountable? – Mys_721tx Oct 28 '18 at 7:54
• @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. – Patrick Stevens Oct 28 '18 at 7:57
• @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 $T$. – Mys_721tx Oct 28 '18 at 22:11