Let $f:R \rightarrow R$, be defined as $f(x)=x^2$.Show that the set$\{x:f(x)<1\}$ is an open set.

Question

Let $f:\mathbb{R} \rightarrow \mathbb{R}$, be defined as $f(x)=x^2$. Show that the set$\{x:f(x)<1\}$ is an open set.

I know how to prove if a set is open or closed. How can I prove if the set is based on a function?

Answer 1

Note that your set $\left\{x:f(x) < 1\right\} = f^{-1}(-\infty,1)$ for any function $f$. If $f$ is continuous, this gives you your answer.

Answer 2

The inverse image under a continuous funtion of an open set is an open set*. The set $X=\{x\in\mathbb{R}:x<1\}$ is open and your set is the inverse image of $X$ and therefore it is open. The proof of * can be found here: Prove that the inverse image of an open set is open

Answer 3

Let $O =\{x:f(x)<1\} $

One way:

• $f$ is continuous. $O =f^{-1}((-\infty,1))$. So, $O$ is the preimage of the open set $(-\infty,1)$ and is open as $f$ is continuous.

Other way:

• $x^2 <1 \Leftrightarrow |x| < 1 \Leftrightarrow -1 < x < 1$. So, $O = (-1,1)$ which is open.

Answer 4

Let $A:=${$x|f(x)<1$}, and

$x_o \in A$, then $f(x_0)<1$.

Since $f(x)$ is continuous at $x_0$:

For $\epsilon >0$ there is a $\delta >0$ s.t.

$|x-x_0| \lt \delta$ implies

$|f(x)-f(x_0)| \lt \epsilon,$ or

$- \epsilon +f(x_0) <f(x)< \epsilon +f(x_0).$

Choose $\epsilon \lt 1-f(x_0) (>0)$.

Then $|x-x_0| \lt \delta$ implies

$f(x) < 1-f(x_0)-f(x_0) =1$.

We found an open ball $B_{\delta}(x_0) \subset A$,

hence $A$ is open.