EnvivyEvoxys6

2022-07-05

Invariance of domain theorem tells us that if a subset $V$ of ${\mathbb{R}}^{n}$ is homeomorphic to an open subset of ${\mathbb{R}}^{n}$, then $V$ must be open itself.

Question: If a subset $V$ of Rn is homeomorphic to a Borel subset of ${\mathbb{R}}^{n}$, must $V$ be Borel ?

Recall Borel(${\mathbb{R}}^{n}$) is defined to be the σ-algebra generated by the topology of ${\mathbb{R}}^{n}$.

Question: If a subset $V$ of Rn is homeomorphic to a Borel subset of ${\mathbb{R}}^{n}$, must $V$ be Borel ?

Recall Borel(${\mathbb{R}}^{n}$) is defined to be the σ-algebra generated by the topology of ${\mathbb{R}}^{n}$.

Tamia Padilla

Beginner2022-07-06Added 16 answers

The answer to the question is yes.

If $B\subseteq {\mathbb{R}}^{n}$ is Borel and $f:{\mathbb{R}}^{n}\to {\mathbb{R}}^{n}$ is continuous such that $f{|}_{B}$ is injective, then the image $f(B)\subseteq {\mathbb{R}}^{n}$ is Borel.

We also have the measurable analog of the Invariance of Domain:

2. If $B\subseteq {\mathbb{R}}^{n}$ is Borel and $f:{\mathbb{R}}^{n}\to {\mathbb{R}}^{n}$ is Borel such that $f{|}_{B}$ is injective, then the image $f(B)\subseteq {\mathbb{R}}^{n}$ is Borel and $f{|}_{B}:B\to f(B)$ is a Borel isomorphism.

(In general one can replace ${\mathbb{R}}^{n}$ with a standard Borel space.)

If $B\subseteq {\mathbb{R}}^{n}$ is Borel and $f:{\mathbb{R}}^{n}\to {\mathbb{R}}^{n}$ is continuous such that $f{|}_{B}$ is injective, then the image $f(B)\subseteq {\mathbb{R}}^{n}$ is Borel.

We also have the measurable analog of the Invariance of Domain:

2. If $B\subseteq {\mathbb{R}}^{n}$ is Borel and $f:{\mathbb{R}}^{n}\to {\mathbb{R}}^{n}$ is Borel such that $f{|}_{B}$ is injective, then the image $f(B)\subseteq {\mathbb{R}}^{n}$ is Borel and $f{|}_{B}:B\to f(B)$ is a Borel isomorphism.

(In general one can replace ${\mathbb{R}}^{n}$ with a standard Borel space.)

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