Timeline for Intersection of zero sets of continuous functions
Current License: CC BY-SA 4.0
7 events
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| Oct 14, 2020 at 12:20 | comment | added | Chris | Thanks! I didn't really get why $m=1$ but then I guess $F$ and $G$ are equivalent to the sets $\{x:\|f(x)\| = 0\}$, $\{x:\|g(x)\| = 0\}$, which are zero-sets of real 1D functions. If $f$, $g$ are close to singularities, what would that imply for $F\cap G$ ? In general, is there any case when the dimension of $F\cap G$ would be $n-1$ ? | |
| Oct 14, 2020 at 11:40 | comment | added | David Roberts♦ | You'd need $f$ and $g$ of maximal rank (eg submersions). Note that $f$ and $g$ define a function $(f,g)\colon \mathbb{R}^n \to \mathbb{R}^{2m}$ and $F\cap G$ is the zero-set of this function. If $F$ and $G$ are $(n-1)$-dimensional, then $m=1$. So I would imagine that away from singularities, $F\cap G$ is $(n-2)$-dimensional. | |
| Oct 14, 2020 at 10:50 | comment | added | Chris | Thanks! I meant Lebesgue measure. Regarding the functions, besides continuity, they are real and analytic. What other assumptions would make answering the question easier ? | |
| Oct 14, 2020 at 10:46 | history | edited | Chris | CC BY-SA 4.0 |
added 9 characters in body
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| Oct 14, 2020 at 10:24 | comment | added | Dieter Kadelka | I think you should edit your question. What are your assumptions about $f$ and $g$? And by measure: Do you mean Hausdorff measure? Lebesgue measure? | |
| Oct 14, 2020 at 9:27 | review | First posts | |||
| Oct 14, 2020 at 9:27 | |||||
| Oct 14, 2020 at 9:26 | history | asked | Chris | CC BY-SA 4.0 |