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 $f(\Bbb R)=\Bbb R$

Can you please help.

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    $\begingroup$ Intermediate value theorem? $\endgroup$ – Angina Seng Dec 5 '18 at 9:07
  • $\begingroup$ @LordSharktheUnknown,Suppose I choose $a\in \Bbb R$ ,I need to show that there exists $b\in \Bbb R$ such that $f(a)=b$ $\endgroup$ – user596656 Dec 5 '18 at 9:13
  • $\begingroup$ How to use IVT here?Can you help $\endgroup$ – user596656 Dec 5 '18 at 9:13
  • $\begingroup$ Why the downvotes?? $\endgroup$ – user596656 Dec 5 '18 at 9:13
  • $\begingroup$ (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". $\endgroup$ – 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 $f(b_k)<r$. We can choose $a_m$ and $b_k$ in such a way that $b_k<a_m$. Now apply the intermediate value theorem to obtain that $f(x)=r$ for some $x\in [b_k,a_m]$.


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