# Proof that $A_c = \{(x_1,x_2) \in \mathbb{R^2}| f(x_1,x_2) < c\}$ is open when $f$ is continuous

I have a proof for the following statement, but I was hoping someone could confirm this for me.

Let $f: \mathbb{R^2} \rightarrow \mathbb{R}$, $c \in \mathbb{R}$, $A_c = \{(x_1,x_2) \in \mathbb{R^2}| f(x_1,x_2) < c\}$. Furthermore, let $f$ be a continous function. Show that $A_c$ is open.

My approach: Let $x=(x_1,x_2) \in A_c$. We know that $f(x_1,x_2) < c$, so we also now that there is a point such that $f(x_1,x_2) < \frac{f(x_1,x_2) + c }{2} < c$.

We also now that $\forall\ y \in \mathbb{R^2}: \forall \epsilon > 0,\ \exists\ \delta > 0: 0 < |x-y| < \delta \implies |f(x_1,x_2) - f(y_1,y_2)| < \epsilon$

Now, let $\epsilon = \frac{c-f(x_1,x_2)}{2} > 0$. Choose $r = \delta$. We know that $B(x,r) = \{ y \in \mathbb{R^2}| |x - y| < r\}$. So, $\forall\ y \in B(x,r)$, it holds that $|x-y|<r=\delta$, which then implies that \begin{equation} |f(x)-f(y)| < \frac{c-f(x_1,x_2)}{2} \end{equation} From which we can conclude that \begin{equation} -\frac{c-f(x_1,x_2)}{2} < f(x) - f(y) \\ \frac{f(x_1,x_2)-c}{2} < \frac{f(x_1,x_2) + c }{2} - f(y) \\ -c < -f(y) \\ f(y) < c\end{equation} So $y \in A_c$, which proves that $A_c$ is open.

• You work too hard. The preimage of an open set by a continuous function is open, and $A_c = f^{-1}(-\infty,c)$. – Alex Provost May 26 '17 at 18:29
• Ah I see. Although, I did not learn that theorem during the lectures, so I doubt that we are allowed to use it though. – user444389 May 26 '17 at 19:01
• What def'n of continuity do you have at your disposal? There are many equivalent def'ns. The most common, general topological def'n is that the inverse of an open set is open. – DanielWainfleet May 26 '17 at 19:08
• It is essentially the proof you gave, in a more general setting. Take a point $x$ in the preimage; since its image $f(x)$ lies in an open set, there is a small ball $B$ around it inside the open set. By continuity of $f$, there is then a small ball around $x$ that is mapped inside $B$; therefore, a fortiori, this ball is also inside the preimage of the open set. – Alex Provost May 26 '17 at 19:16
• I used the following definition: $\forall x,y \in D: \forall\ \epsilon > 0 \exists\ \delta > 0: 0 < |x-y|, \delta \implies |f(x)-f(y)| < \epsilon$. – user444389 May 26 '17 at 20:07