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


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


$T$ is uncountable.


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


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


$T$ is uncountable.


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

  • 4
    $\begingroup$ $\Bbb{Z}$ 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

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.

  • 2
    $\begingroup$ 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
  • 1
    $\begingroup$ @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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.