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Let $$f(x) = \dfrac{x}{1+x^2}.$$ I am trying to determine if the image $f([0,2])$ is closed or not closed. I think it's not closed but I don't know how to show it.

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Hint: The continuous image of a compact set is compact.

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up vote 2 down vote accepted

Note that for $x > 0$, we have $$x + \dfrac1x \geq 2 \implies \dfrac{x^2+1}x \geq 2 \implies \dfrac{x}{1+x^2} \leq \dfrac12$$ The equality holding when $x=1$, which is in the interval $[0,2]$. Also, $\dfrac{x}{x^2+1} \geq 0$ for $x \in [0,2]$ and this minimum is hit for $x=0$. Hence, the image is the interval $$\left[0,\dfrac12 \right]$$

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So are there any closed sets that get mapped to open sets? – user39794 May 11 '13 at 3:11
Closed set $[1,+\infty)$ is mapped to a non-closed set by this function. – GEdgar May 11 '13 at 13:22

Another way to do it without AM-GM inequality:

Let $ x = \tan \theta $ and hence $ \frac{x}{x^2 + 1} = \frac{\tan \theta}{\sec^2 \theta} = \frac{1}{2} \sin 2 \theta $, which is easily seen to have an upper bound of $ \frac{1}{2} $.

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Another way to prove that the continuous image of a closed interval is again a closed interval is to apply the intermediate value theorem (which immediately implies that the image is again an interval) together with the extreme value theorem (which immediately applies that the image contains the endpoints). This is probably more natural in the calculus context than an appeal to compactness.

Also, you can ask WolframAlpha what the image of a set looks like: "image of x/(1+x^2) restricted to [0,2]"

enter image description here

The result of $[0,1/2]$ agrees with one of the other answers.

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HINT: Closed.

If you observe the function $f(x)$ starts from 0 at $x=0$, increases to a max. point and the decreases till it reaches $x=2$. Let $m$ be the max. value which you can calculate.

Clearly, [0,m] is closed.

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