# Inverse image of an open set with a function

Let $$f: (\mathbb{R},|.|)\to (\mathbb{R},|.|)$$ defined by $$f(x)=\dfrac{1}{1+x^2}$$

I want to find $$f^{-1}(]a,b[)$$

What I do: $$f^{-1}(]a,b[)=\{y\in \mathbb{R}, \frac{1}{1+y^2}\in ]a,b[\}$$

that is : $$a<\frac{1}{1+y^2}

• If $$a>0$$ then $$\frac1b-1 after that we must suppose two sases $$\frac1b\leq 1$$ or $$\frac1b>1$$

## Is there an other method ?

i suppose that $$a $$f^{-1}(]a,b[)= \begin{cases} \emptyset,~ b<0\\ ]-\infty,-\sqrt{\frac1b-1}[\cup]\sqrt{\frac1b-1},+\infty[,~ a\leq0 ~\text{and}~b<1\\ \mathbb{R},~ b>1,a\leq0\\ \mathbb{R}\setminus{0},~ b=1,a\leq0\\ \emptyset,~b\geq1,a>1\\ ]-\sqrt{\frac1a-1}, \sqrt{\frac1a-1}[,~b\geq1,a<1\\ ]\sqrt{\frac1b-1},\sqrt{\frac1a-1}[,~ b<1, a>0 \end{cases}$$

Thank you

• You are on the right track. You laso have to consider the case $a<0$ and $b>0$. Mar 1, 2019 at 10:24
• There are three cases to consider $a > 1, a \leq 1 < b$ and $a < b \leq 1.$ Mar 1, 2019 at 10:28
• @KaviRamaMurthy please see my edits and tel me if it is correct
– user523857
Mar 1, 2019 at 10:56
• @Dbchatto67 please see my edits and tel me if it is correct
– user523857
Mar 1, 2019 at 10:57

I think a good way to approach the problem is to think first about the function $$f$$:

First of all, since $$x^{2} \geq 0$$ then $$1+x^{2} \geq 1$$ which leads to $$f(x) = \frac{1}{1+x^{2}} \leq 1$$. This means that if $$a > 1$$ then $$f^{-1}(]a,b[) = \emptyset$$.

We also have that $$1+x^{2} \geq 1 > 0$$ so $$f(x) = \frac{1}{1+x^{2}} >0$$. This means that if $$b \leq 0$$ then we also have $$f^{-1}(]a,b[) = \emptyset$$.

Furthermore, $$\forall a,a' \in ]\infty, 0]$$ and $$b>0$$ we have $$f^{-1}(]a,b[) = f^{-1}(]a',b[)$$ and $$\forall a < 1$$ and $$b \in ]1, \infty]$$ we have $$f^{-1}(]a,b[) = f^{-1}(]a,b'[)$$

With this results, we can then separate the problem in appropriate cases:

$$\underline{\text{Case} \,\, 1:}$$ $$0

$$a < f(x) < b \,\,\Leftrightarrow \,\, a < \frac{1}{1+x^{2}} < b \,\,\overset{a,b >0}{\Longleftrightarrow}\,\, \frac{1}{b} < 1+x^{2} < \frac{1}{a} \,\,\Leftrightarrow \,\,\frac{1}{b}-1 < x^{2} < \frac{1}{a}-1$$

Now, since $$a, b\leq 1$$, we have that $$\frac{1}{b}-1 \geq 0$$ and $$\frac{1}{a}-1 \geq 0$$ and we can take square roots to have

$$\frac{1}{b}-1 < x^{2} < \frac{1}{a}-1 \,\, \Leftrightarrow \,\, \sqrt{\frac{1}{b}-1} < |x| < \sqrt{\frac{1}{a}-1}$$

(notice that the fact of the square root being an increasing injective function allows us to take square roots and not changing the grater-than and less-than symbols and to have the equivalence symbol)

We finally have that $$x \in f^{-1}(]a,b[) \,\,\Leftrightarrow \,\, x\in \,\, ]\sqrt{\frac{1}{b}-1},\sqrt{\frac{1}{a}-1}[ \quad \text{or} \quad x\in \,\, ]-\sqrt{\frac{1}{a}-1},-\sqrt{\frac{1}{b}-1}[$$

We put now our attention on the limit cases:

$$\underline{\text{Case} \,\, 2:}$$
$$\underline{\text{Subcase} \,\, 1:}$$ $$a=0$$ and $$b \leq 1$$

$$a=0 < \frac{1}{1+x^{2}}$$ so the condition $$a is equivalent to $$f(x).

$$\underline{\text{Subcase} \,\, 2:}$$ $$a>0$$ and $$b > 1$$

We have $$\frac{1}{1+x^{2}} \leq 1 so the condition $$a is equivalent to $$a.

We can summarize the problem in the next cases:

The empty cases:

$$a > 1\, \Rightarrow \,f^{-1}(]a,b[) = \emptyset$$

$$b \leq 0 \, \Rightarrow \,f^{-1}(]a,b[) = \emptyset$$

The whole case:

$$a \leq 0$$ and $$b >1$$ $$\, \Rightarrow \,f^{-1}(]a,b[) = \mathbb{R}$$

The partial cases:

$$a \leq 0$$ and $$0 < b \leq 1$$ $$\, \Rightarrow \,f^{-1}(]a,b[) = ]-\infty,-\sqrt{\frac{1}{b}-1}[ \,\, \bigcup \,\, ]\sqrt{\frac{1}{b}-1},\infty[$$

$$1 > a > 0$$ and $$b > 1$$ $$\, \Rightarrow \,f^{-1}(]a,b[) = ]-\sqrt{\frac{1}{a}-1},\sqrt{\frac{1}{a}-1}[$$

The restrictive case:

$$0 and $$0 < b \leq 1$$ $$\, \Rightarrow \,f^{-1}(]a,b[) = ]-\sqrt{\frac{1}{a}-1},-\sqrt{\frac{1}{b}-1}[ \,\, \bigcup \,\, ]\sqrt{\frac{1}{b}-1},\sqrt{\frac{1}{a}-1}[$$

Notice that the good choice of the limit points ($$0$$ for $$a$$ and $$1$$ for $$b$$) is the key.

The proof may appear very long but if you draw the function (maybe with WolframAlpha) you will understand it much better.

I hope it helps you!

• please if I want to prove that $]x-r,x+r[\subset f^{-1}(]x-\varepsilon, x+\varepsilon[)$ without find exactly the pre image how to do please?
– user523857
Apr 13, 2019 at 19:46
• This is not true in general... What do you mean exactly? prove the continuity of the function by definition? Apr 14, 2019 at 20:36