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I am working through my study guide for my calculus class for the test we're having tomorrow. There are two problems that I'm not completely sure about.

For each of the following statements, determine whether it is true or false and justify your answer.

  1. For any function $f:\left[0,1\right]\rightarrow\mathbb{R}$, its image $f\left(\left[0,1\right]\right)$ is an interval.
  2. For any continuous function $f:D\rightarrow\mathbb{R}$, its image $f\left(D\right)$ is an interval.

For the first one, I said:

False. Let $f(x)=\frac{1}{x-\frac{1}{2}}$, $f$ is not continuous at $x=\frac{1}{2}$, and thus the image is not an interval.

For the second one, I said:

False. If $D$ is not an interval.

For the first one, I am decently confident that my answer is correct. I would just like to know what you think, if it is sufficient enough.

However, for the second one, I don't know where to go for the second one. The professor would like an example, so maybe something like $D=(-\infty,0)\cup(0,\infty)$ and $f(x)=\frac{1}{x}$? Because, then the image of $f(D)$ would not include 0, and thus not be an interval. Would that work as an example?

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    I think it's more helpful to learn to find easiest examples. You have one for 2 (Stefan above gives you the technical reason why it works, or can work); but your example for 1 (which can be made ok) is much too difficult. Recall that a function is just any functional relation between 2 spaces. Try something constant first. Here, mapping 0 to 1, and all other$x$in (0, 1] to 0 is a very easy counter you'll instantly see works.2012-12-02

1 Answers 1

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Here are some easy counterexamples:

  1. Define $f(x) := \begin{cases} 0 & x \in \left[0,\frac{1}{2}\right] \\ 1 & x \in \left(\frac{1}{2},1 \right] \end{cases}$ Then $f([0,1]) = \{0,1\}$ and this is obviously no interval.
  2. You already mentioned $g(x) := \frac{1}{x}$ for $x \in D:=(-\infty,0) \cup (0,\infty)$. That works. Another (maybe easier) one would be $h(x) := x \qquad \qquad D=:[0,1] \cup [2,3]$ Then $h(D)=[0,1] \cup [2,3]$ and this is -again- no interval (but $h$ is continuous).