- #1
Eclair_de_XII
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- 91
- TL;DR
- Assume that I've only taken math courses up until Calculus III. Informal definition is as follows:
Let ##f:\mathbb{R}^2\longrightarrow\mathbb{R}## be defined. Then the level set of ##f## w.r.t. some point ##z\in f(\mathbb{R}^2)## is the set of points ##(x,y)## in the ##xy##-plane with the property that ##f(x,y)=z##.
Let ##f,g:\mathbb{R}^2\longrightarrow\mathbb{R}## be defined, and denote ##D=f(\mathbb{R}^2)##. Assume without loss of generality that ##g(\mathbb{R}^2)\equiv f(\mathbb{R}^2)##.
Define a function ##\varphi_f:D\longrightarrow \mathbb{R}^2## as follows: ##\varphi_f(z)=\{(x,y):f(x,y)=z\}##, and define ##\varphi_g## similarly. Let ##z\in D##. Suppose that ##\varphi_g\equiv\varphi_f##. Then ##\{(x,y):f(x,y)=z\}=\{(x,y):g(x,y)=z\}##. In other words, the point ##z## associates the functions ##f,g## to the same exact points ##(x,y)## in ##\mathbb{R}^2##.
Hence, for every ##(x,y)\in \varphi_f(z)##:
\begin{eqnarray*}
(x,y,f(x,y))=(x,y,g(x,y))\\
(0,0,(f-g)(x,y))=(0,0,0)
\end{eqnarray*}
It follows then, that ##f\equiv g##.
This isn't actually a proof; I'm just asking a question that I cannot find an answer to on the internet. This is just a rudimentary explanation of why I think level sets are unique in ##\mathbb{R}^2##.
Define a function ##\varphi_f:D\longrightarrow \mathbb{R}^2## as follows: ##\varphi_f(z)=\{(x,y):f(x,y)=z\}##, and define ##\varphi_g## similarly. Let ##z\in D##. Suppose that ##\varphi_g\equiv\varphi_f##. Then ##\{(x,y):f(x,y)=z\}=\{(x,y):g(x,y)=z\}##. In other words, the point ##z## associates the functions ##f,g## to the same exact points ##(x,y)## in ##\mathbb{R}^2##.
Hence, for every ##(x,y)\in \varphi_f(z)##:
\begin{eqnarray*}
(x,y,f(x,y))=(x,y,g(x,y))\\
(0,0,(f-g)(x,y))=(0,0,0)
\end{eqnarray*}
It follows then, that ##f\equiv g##.
This isn't actually a proof; I'm just asking a question that I cannot find an answer to on the internet. This is just a rudimentary explanation of why I think level sets are unique in ##\mathbb{R}^2##.
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