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Let $X:B_1(0)\to \mathbb{R}^3 \in C^2(B_1(0), \mathbb{R}^3)$ be a mapping from $B_1(0)\subset \mathbb{R}^2$ into $\mathbb{R}^3$. Is it true that $X(B_1(0))$ doesn't contain any boundary points? I don't have a concrete definition, but what I mean by boundary point is: https://math.stackexchange.com/a/2302147/776794

It might be relevant that $X$ is a (not necessarily regular) minimal surface, which means

$$\lvert X_u\rvert^2 = \lvert X_v\rvert^2,\,\,\,\, \langle X_u, X_v\rangle = 0,\,\,\,\, \Delta X = 0 \,\,\text{ on } B_1(0)$$

Mandelbrot
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  • What is $\Omega$? The domain of $X$ is the ball. – Ted Shifrin Dec 31 '20 at 18:22
  • Sorry, what I meant was $B_1(0)$. – Mandelbrot Dec 31 '20 at 18:24
  • Maybe it's helpful to note that non-constant harmonic functions are open maps. – Ted Shifrin Dec 31 '20 at 18:35
  • I was thinking about something slightly similar. Since $B_1(0)$ is simply connected, a theorem from the book I am reading tells us that there exists a holomorphic function $f:B_1(0)\to \mathbb{C}^3$ such that $\Re(f)=X$. Since $f$ is holomorphic and by that an open map, can we conclude that $X$ doesn't have any boundary points? – Mandelbrot Dec 31 '20 at 18:40
  • For every $x\in f(B_1(0))$, there is an $\epsilon>0$ such that $B_{\epsilon}(x)\subset f(B_1(0))$. If we restrict $f$ to its real part, we should still be able to find open neighbourhoods of every component of $\Re(x)$, which means that $x$ can't be a boundary point. Is this correct? – Mandelbrot Dec 31 '20 at 18:45
  • You have to formalize boundary point a bit more, but think about the boundary of a half-plane as the zero set of a real coordinate function. – Ted Shifrin Dec 31 '20 at 18:48
  • I agree, but this concept wasn't formally introduced in my book and it isn't necessary for the any proof or the presentation I am giving. I am just looking for a vivid explanation in the case that someone asks a question. – Mandelbrot Dec 31 '20 at 18:50

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