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I am stuck on the following problem:

Let $f :\mathbb R \to \mathbb R$ be a function such that $f$ is not bounded above nor bounded below. Show that if $f$ is continuous on $\mathbb R$ then the image of $f$ is $\mathbb R$.

I'm not sure how to exactly approach this problem, especially because it's so intuitive. I know you have to use the Intermediate Value Theorem.

5xum
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jeff123
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  • Welcome to MSE. Please include your question in the body of the question, instead of putting it only in the title. – José Carlos Santos Feb 04 '20 at 08:46
  • Welcome to MSE, please note that I edited your question to fit in with our standards. In parituclar, I used mathjax to format it, and reformatted it so the question is not in the title, but the body of your post. In future questions, please keep this in mind. Do not expect community members to always clean up your questions the way this one was cleaned. – 5xum Feb 04 '20 at 08:48

3 Answers3

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Hint: for each $M>0$, there exist $x_1, x_2$ such that $f(x_1)>M$ and $f(x_2) < -M$. What does that tell you about $[-M, M]$?

5xum
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Note that for every natural number $n$ $f$ must attain a value $y_n>n$ since otherwise $f$ would be bounded. By the IVT $f$ attains each value between $y_n$ and $y_{n+1}$ for any $n$. So $f$ is surjective onto $[m,\infty)$, where $m=\min_n y_n$. Doing the same for negative values and filling the gap in between, we conclude that $f$ has range $\mathbb R$.

Jonas Linssen
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Use IVT : Let's assume a real number $M$ such that $f(x)\neq M$ for all $x$, but by the unboundedness property of the function there is $x_1$ with $f(x_1)>M$ and $f(x_2)<M$, now using IVT $f$ is gonna attain all the values between $f(x_1), f(x_2)$, so it would attain $M$ too since $f(x_2)<M<f(x_1)$. So the assumtion was not true.

aud098
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