# Can any bounded area defined by polynomial inequality in $\mathbb{R}^n$ be partitioned into simply connected finite area such that

Can any bounded area defined by polynomial inequality in $\mathbb{R}^n$ be partitioned into simply connected finite areas such that for each simply finite area there exist a diffeomorphic map that maps the area to a m-sphere? Bounded means the area $A<\infty$ with the boundary defined by the polynomials. To partition means to partition with k-dimension manifold($k=m-1$) as boundary(the intersection of the two subset) .

• What does "partitioned" mean? A disjoint union? (This clearly can't happen.) An open cover? – Qiaochu Yuan Nov 15 '14 at 6:43
• @QiaochuYuan, thank you, I will edit it to clarify the question. – XL _At_Here_There Nov 15 '14 at 6:53
• I still don't understand the question. The $n$-sphere cannot even immerse into $\mathbb{R}^n$. Did you mean the $n-1$-sphere, or maybe the $n$-disk? – Qiaochu Yuan Nov 15 '14 at 7:17
• @QiaochuYuan, thank you for your patience. Let me give a example in 1-dimension, $\int_a^b f(x)dx=\int_a^c f(x)dx+\int_c^b f(x)dx$, [a,b] is partitioned into [a,c],[c,b]. – XL _At_Here_There Nov 15 '14 at 7:25
• it is well-known that any semialgebraic set can be triangulated. The question you ask seems like a particular case of this fact. – Dima Pasechnik Nov 15 '14 at 9:41