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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.

Any comments?

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  • $\begingroup$ You work too hard. The preimage of an open set by a continuous function is open, and $A_c = f^{-1}(-\infty,c)$. $\endgroup$ – Alex Provost May 26 '17 at 18:29
  • $\begingroup$ Ah I see. Although, I did not learn that theorem during the lectures, so I doubt that we are allowed to use it though. $\endgroup$ – user444389 May 26 '17 at 19:01
  • $\begingroup$ 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. $\endgroup$ – DanielWainfleet May 26 '17 at 19:08
  • $\begingroup$ 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. $\endgroup$ – Alex Provost May 26 '17 at 19:16
  • $\begingroup$ I used the following definition: $\forall x,y \in D: \forall\ \epsilon > 0 \exists\ \delta > 0: 0 < |x-y|, \delta \implies |f(x)-f(y)| < \epsilon$. $\endgroup$ – user444389 May 26 '17 at 20:07
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Your reasoning is rigorous and correct. Besides minor things like writing style that may admit a room for discussion, I read and followed your proof easily. In fact, among the similar questions here I have ever seen your work as I see it is the nicest one; you know what you are talking about.

By the way, as a commenter has pointed out, sometimes you may want to make some preliminary observations on the given problems first in order to solve it economically and elegantly. You know, "blindly" "overworking" every given problem without seeing the whole picture in the first place may not be a good habit in the long run. But I guess it is no doubt that this "overworking" habit is good for self-training.

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  • $\begingroup$ Thanks! During the lectures, I was not given the theorem that the preimage of an open set by a continous function is open, so I did not use that. This seemed to be the easiest way whithout that theorem. $\endgroup$ – user444389 May 26 '17 at 18:51
  • $\begingroup$ No problem. I hoped to be more helpful; so I mentioned it just in case you are not aware of it. All the best :). $\endgroup$ – Megadeth May 26 '17 at 18:54

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