# Strictly increasing and continuous function

I have to prove that if $$f\to \infty$$ in a stricly increasing way and f is continuous then if $$y_0>f(x_0)$$, $$\exists !\, x_1>x_0$$ such that $$f(x_1)=y_0$$.
My idea is that if f is continuous then it takes all values greater than $$f(x_0)$$ one only time, but how could I formalize this?

• Are you allowed to use Bolzano's theorem? Commented Jan 11, 2023 at 9:54
• @A.J.Pan-Collantes this can help me to say that f takes all values up to $\infty$, but why one only time?
– axi
Commented Jan 11, 2023 at 9:56
• As written, the statement is wrong. It should be: if $y_0>f(x_0)$, then $\exists !\, x_1>x_0$ such that $f(x_1)=y_0$. ($x_1$ will depend on $y_0$). Commented Jan 11, 2023 at 9:58
• That is because it is strictly increasing Commented Jan 11, 2023 at 9:58
• @geetha290krm me too. Maybe I had written not so well the question...
– axi
Commented Jan 11, 2023 at 10:00

Since $$f(x) \to \infty$$ as $$x \to \infty$$, there exists $$z>x_0$$ such that $$f(z)>y_0$$. Apply the intermidiate value theorem to $$f$$ on $$[x_0,z]$$. Since $$y_0$$ lies between $$f(x_0)$$ and $$f(z)$$ there exists $$x_1$$ with $$f(x_1)=y_0$$. $$x_1$$ is unique because $$f$$ is strictly increasing.
[If $$f(x_1)=f(x_2)$$ then we can neither have $$x_1 nor $$x_2 by strict monotonicity].