# Show that $f(\Bbb R)=\Bbb R$.

Let $$f$$ be a continuous function.Suppose there exists a sequence of points $$a_n\to \infty$$ such that $$f(a_n)\to \infty$$ and there exists a sequence of points $$b_n\to -\infty$$ such that $$f(a_n)\to -\infty$$.

Show that $$f(\Bbb R)=\Bbb R$$.

Given any $$M>0$$ there exists $$n_1\in \Bbb N$$ such that $$f(a_n)>M\forall n>n_1$$ .

Similarly there exists $$n_2\in \Bbb N$$ such that $$f(a_n)<-M\forall n>n_2$$ .

I am unable to show that $$f(\Bbb R)=\Bbb R$$

• @LordSharktheUnknown,Suppose I choose $a\in \Bbb R$ ,I need to show that there exists $b\in \Bbb R$ such that $f(a)=b$ – user596656 Dec 5 '18 at 9:13
• (i) so far there is only one downvote. (ii) The IVT can be rephrased as "let $f$ be a continuous function $I\to\Bbb R$ where $I$ is an interval, then $f(I)$ is an interval". – Angina Seng Dec 5 '18 at 9:16
Suppose $$r\in\mathbb{R}$$. We want to show that $$f(x)=r$$ for some $$x\in\mathbb{R}$$.
Since $$f(a_n)\to\infty$$ then there exists $$m$$ such that $$f(a_m)>r$$. Analogously $$f(b_n)\to -\infty$$ and so there exists $$k$$ such that $$f(b_k). We can choose $$a_m$$ and $$b_k$$ in such a way that $$b_k. Now apply the intermediate value theorem to obtain that $$f(x)=r$$ for some $$x\in [b_k,a_m]$$.