let g rn be open and f g rm be continuously differentiable such that df x has full rank for all

Name: let g rn be open and f g rm be continuously differentiable such that df x has full rank for all x g show that f g is open if m n 51572
Uploaded: 2023-07-17T19:18:13-08:00
Duration: 6 min 6 s
Channel: Madhur L
Description: let g rn be open and f g rm be continuously differentiable such that df x has full rank for all x g show that f g is open if m n 51572

Step 1: Since G is open, for any point x in G, there exists an open ball B(x, r) contained in G.

Let g rn be open and f g rm be continuously differentiable...

Transcript

00:01 Hello everyone, welcome to this video.

00:03 Here we are given with f from r to r be a function and we have to prove that if f is continuous then f inverse of g is open in r whenever g is open in r.

00:14 So first let us prove by taking that f is continuous on r.

00:23 So if f is continuous on r then there will be an element a which belongs to f inverse of g.

00:31 So by definition we can write f of a belongs to g.

00:37 Since g is open then there exists a epsilon greater than zero such that the ball of f of a comma epsilon will be contained in g.

00:59 So since f is continuous at a then there exists delta greater than zero such that x minus a modulus less than delta which implies that f of x minus f of a will be less than epsilon.

01:28 So that is x belongs to b of a comma delta which implies f of x belongs to b of f of a comma epsilon.

01:44 So in particular we can say f of x belongs to capital g.

01:53 So therefore we can say b of a comma delta is contained in f inverse of g.

02:01 So which means that f inverse of g is open.

02:08 Now we shall prove the part.

02:13 So suppose here let us take f inverse of g is open then we have to say that f is continuous.

02:23 So to prove here that is f is continuous.

02:29 Now here let us take a belongs to r where epsilon greater than zero and we know that a ball that is b of f of a comma epsilon is open in r then we can say f inverse of b of f of a comma epsilon as a belongs to as you can write it here a belongs to f inverse of b of f of a comma epsilon.

03:19 So which implies delta greater than zero such that b of a comma delta is contained in f inverse of b of f of a comma epsilon.

03:36 Now from this we can write that x minus a modulus is less than delta which implies that f of x minus f of a will be less than epsilon.

03:53 So from this we can say that by our definition we can say that f is continuous continuous at a.

04:07 So hence we proved the theorem...

Question

Let G ⊆ Rn be open and f : G → Rm be continuously differentiable such that [Df (x)] has full rank for all x ∈ G. Show that f (G) is open, if m ≤ n.

Question

Let G ⊆ Rn be open and f : G → Rm be continuously differentiable such that [Df (x)] has full rank for all x ∈ G. Show that f (G) is open, if m ≤ n.

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