The natural boundary of the modular $\lambda$ function is the real axis I want to solve the following problem from Ahlfors' text:

Show that the function $\lambda(\tau)$ introduced in Chap. 7, Sec. 3.4, has the real axis as a natural boundary.

Here $\lambda(\tau)$ is the modular lambda function, whose definition in the text appears in a previous question of mine.
My attempt:
From the definition $$\lambda(\tau)= \frac{e_3-e_2}{e_1-e_2} $$ It can be seen that the numerator and the the denominator, which are both analytic in the upper half plane $\Im \tau >0$, have the real axis as their natural boundary. This is because, as mentioned in Daniel Fischer's answer, they both have a dense subset of poles on the real line. The numerator has poles in all points of the form $$\frac{1-2n_1}{-1+2n_2}  $$ and $$\frac{2n_1}{1-2n_2}, $$ where $(n_1,n_2) \in \mathbb Z^2 \setminus \{(0,0 \}$ . Similarly the denominator has poles at the points $$\frac{1-2n_1}{2n_2} $$
and $$\frac{2n_1}{1-2n_2} $$.
Note that their sets of poles intersect, and only the latter points remain poles at the quotient $\lambda$. I would like to say that the remaining set of poles exclusive for the numerator is also dense, and when approaching any such point the denominator doesn't grow too large. Unfortunately, it appears that I can always find poles of the denominator which get arbitrarily close to any pole of the numerator, which is giving me a hard time.
Is my reasoning correct? If so, please help me overcome the problem of denseness of the poles of the denominator. If not, please suggest an alternative way to approach this problem.
Thank you.
 A: I would look at the mapping properties. $\lambda$ maps the $0$-angled (hyperbolic) triangle $\Delta$ with vertices $0,1,\infty$ conformally onto the upper half plane. That means it stretches a $0$-angle to an angle $\pi$. The same happens at the vertices of all (iterated) reflections of $\Delta$, and every point of $\mathbb{R}$ is an accumulation point of vertices of reflections of $\Delta$. But a holomorphic function cannot stretch a $0$-angle to a positive angle.
Regarding your strategy,

I would like to say that the remaining set of poles exclusive for the numerator is also dense, and when approaching any such point the denominator doesn't grow too large.

is almost true, every rational number with odd numerator and denominator is a "pole" for the numerator, but not for the denominator, and that set of rationals is dense.
However, these are not poles. Every one of these points is an accumulation point of points that $\lambda$ maps to $\infty$ (extending $\lambda$ by continuity to the closure of $\Delta$, and by reflection to the vertices of the triangles on $\mathbb{R}$); and of points that $\lambda$ maps to $0$, and of points that $\lambda$ maps to $1$. And as a consequence, there are sequences $(z_n)$ in the upper half plane converging to such a point with $\lambda(z_n) \to \infty$, and sequences $(w_n)$ converging to the same point with $\lambda(w_n) \to 0$, which shows that $\lambda$ cannot be analytically continued across any part of $\mathbb{R}$. But while the mapping properties of $\lambda$ yield that rather directly, I don't see an easy way of proving that by considering the numerator and denominator separately.
