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  1. Find a continuous function on a closed interval with range an open interval.

I'm having trouble thinking of an example of such a function. I though $f(x)=x$ from $\mathbf{R}$ to $\mathbf{R}$ would work, but $\mathbf{R}$ isn't an interval. Is my idea not too far off, i.e., is there a slight modification that will work to deal with the intervals?

  1. Is there a continuous function defined on an open interval with range an unbounded closed set different from $\mathbf{R}$? I think that there is but I'm not sure, I really would prefer to not get an answer, just a hint.
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You seem to be excluding unbounded intervals like $(-\infty, +\infty)$ or $[0, +\infty)$ from your definition of interval. If not see Travis's answer. If so:

For 1, a continuous image of the (bounded) closed interval $[x, y]$ must be compact and so cannot be an open interval $(a, b)$ (unless $x > y$ so that $[x, y] = \emptyset$).

As a hint for 2, think of a function $f$ that maps the open interval $(-1, 1)$ onto the unbounded closed set $[1, \infty)$ (which is ${} \neq \Bbb{R}$) with $f(0) = 1$ and with $f(x)$ increasing to $\pm\infty$ as $x$ tends from $0$ to $\pm1$. Now see if you can find a quadratic that looks like the reciprocal of the function you're thinking of.

$f(x) = \frac{1}{1-x^2}$

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  1. The real line, $\Bbb R$, is certainly an open interval. In particular, the identity function $f(x) := x$ satisfies the condition. (In fact, for any finite, closed interval $[a, b]$ and continuous function $f$, $[a, b]$ is compact and so $f([a, b])$ is compact and nonempty and hence not open. So, for any continuous $f$ and interval $I$ satisfying the criteria, we must have that $I$ is infinite, which here I mean to include half-infinite.)

  2. Hint Can you think of a familiar continuous function $f$ that has domain $\Bbb R$ but range $[0, \infty)$?

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  • $\begingroup$ For 2, is $1/x^2$ getting me on the right track? I know I'm missing 0 in the domain $\endgroup$ – Guest23 Oct 20 '15 at 23:25
  • $\begingroup$ @Guest23: Travis is probably thinking of $\exp$ but it doesn't meet your requirements because its range is not a closed subset of $\Bbb{R}$. $\endgroup$ – Rob Arthan Oct 20 '15 at 23:52
  • $\begingroup$ @RobArthan Oops, of course I meant the closed interval. There are a few simple examples that fit the (corrected) hint. $\endgroup$ – Travis Oct 21 '15 at 0:09

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