- #1
Saketh
- 261
- 2
Homework Statement
For any open set [itex]U \subset \mathbb{R}^n[/itex] and any continuous and injective mapping [itex]f : U \rightarrow \mathbb{R}^n[/itex], the image [itex]f(U)[/itex] is open, and [itex]f(U)[/itex] is a homeomorphism.
Homework Equations
N/A
The Attempt at a Solution
I am trying to learn how to write proofs, so my logic might be flawed.
Lemma: f is bijective.
Proof: The cardinality of U and [itex]\mathbb{R}^n[/itex] is the same. Because f is one-to-one, all that remains is to prove that every image has a corresponding preimage, which follows directly from the continuity of f.
f(U) is necessarily open because of the definition of continuity.
The inverse of f exists and is continuous because f is bijective and continuous (I feel like that statement is invalid). Therefore, f is homeomorphic.