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Thread: Proof that the range of f is all of R

  1. #1
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    Proof that the range of f is all of R

    Hey all, I need some help with this proof for my analysis class. I have a proof that I wrote up but I'm not sure if it's correct. I'll write the claim and my idea of a proof below and if anyone could tell me what's wrong with my logic or if there's a better way to write the proof, I'd really appreciate it. I'm just not sure how to go about this so what I'm providing is my best guess.

    Claim: Assume $\displaystyle f$ is a continuous real-valued function with domain $\displaystyle (-\infty , \infty )$. Assume $\displaystyle \lim_{x \rightarrow \infty} f(x) = \infty$ and $\displaystyle \lim_{x \rightarrow - \infty} f(x) = - \infty$. Prove that the range of $\displaystyle f$ is $\displaystyle (-\infty , \infty )$.

    Proof: Assume the hypotheses. Since f is continuous at every point in its domain, we have that $\displaystyle \lim_{x \rightarrow x_0} f(x) = f(x_0)$ for all $\displaystyle x_0 \in (-\infty , \infty )$. Then by the definition of limit we have that for all $\displaystyle \varepsilon$ > 0, there exists a $\displaystyle \delta$ > 0 such that $\displaystyle |f(x) - f(x_0)| < \varepsilon$ whenever $\displaystyle |x - x_0| < \delta$. That is, $\displaystyle f(x_0) - \varepsilon < f(x) < f(x_0) + \varepsilon$. So if we let $\displaystyle x_0 \rightarrow \infty$, then $\displaystyle f(x_0) \rightarrow \infty$ and so $\displaystyle f(x) < \infty$. Also, if we let $\displaystyle x_0 \rightarrow -\infty$, then $\displaystyle f(x_0) \rightarrow -\infty$ and so $\displaystyle -\infty < f(x)$. Then $\displaystyle -\infty < f(x) < \infty$, so the range of f is $\displaystyle (-\infty , \infty )$.

    Again, I know it seems sketchy, so please help me identify what I need to say to make this more precise.
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  2. #2
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    Well, how about this:

    Take $\displaystyle n\in \mathbb{N}$ and $\displaystyle N_1,N_2$ such that if $\displaystyle x<N_1$ then $\displaystyle f(x)\leq -n$ and if $\displaystyle x>N_2$ then $\displaystyle f(x)\geq n$ (these $\displaystyle N$'s exist by your limit conditions) then by the Intermediate value theorem we have that for every $\displaystyle y\in (-n,n)$ there exists a $\displaystyle z\in \mathbb{R}$ such that $\displaystyle f(z)=y$. Since this is true for any $\displaystyle n$ we get the result.
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  3. #3
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    Quote Originally Posted by Jose27 View Post
    Well, how about this:

    Take $\displaystyle n\in \mathbb{N}$ and $\displaystyle N_1,N_2$ such that if $\displaystyle x<N_1$ then $\displaystyle f(x)\leq -n$ and if $\displaystyle x>N_2$ then $\displaystyle f(x)\geq n$ (these $\displaystyle N$'s exist by your limit conditions) then by the Intermediate value theorem we have that for every $\displaystyle y\in (-n,n)$ there exists a $\displaystyle z\in \mathbb{R}$ such that $\displaystyle f(z)=y$. Since this is true for any $\displaystyle n$ we get the result.
    Wow, that works very nicely. And it makes a lot more sense that I would need to use the IVT, since we just learned about it not too long ago. Thanks for your help!
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