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Let $M\subseteq\mathbb{R}^{2+n}$ be open, $f\colon M\to\mathbb{R}^n$ be continious. Furthermore consider $A\subseteq\mathbb{R}^{1+n}, A\subseteq M$. My questions are if then (1) $A$ is open and (2) $f$, restricted to $A$, is continious.

(2) I think thats right, because continious on $M$ means continious in every point of $M$. And because $A\subseteq M$ this holds for every point in $A$.

But with (1) I do not know how to answer.

Asaf Karagila
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  • Since $A\subseteq \mathbb R^{1+n}$ as well as $A\subseteq M\subseteq \mathbb R^{2+n}$ and $\mathbb R^{1+n}$ is disjoint from $\mathbb R^{2+n}$, we must have $A=\varnothing$, which is open. – hmakholm left over Monica May 02 '14 at 12:28
  • Ehm, why is $\mathbb{R}^{1+n}$ disjoint from $\mathbb{R}^{2+n}$? Thats not true. –  May 02 '14 at 12:31
  • x @math12: Each element of $\mathbb R^{1+n}$ is a tuple of $1+n$ reals. Each element of $\mathbb R^{2+n}$ is a tuple of $2+n$ reals. Nothing is simultaneously an $(1+n)$-tuple and a $(2+n)$-tuple, so $\mathbb R^{1+n}$ and $\mathbb R^{2+n}$ have no elements in common -- in other words, they are disjoint. – hmakholm left over Monica May 02 '14 at 13:08

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