I want to prove that for a continuous function mapping a connected space to ℝ such that f(p) never e

lifretatox8n

lifretatox8n

Answered question

2022-05-10

I want to prove that for a continuous function mapping a connected space to ℝ such that f(p) never equals s, it follows that f(p) < s for all p or f(p) > s for all s.
So here's what I know so far:
-Because f is continuous and the metric space M is connected, f(M) is connected.
-f(M) is connected in the real numbers, so if x, y ϵ f ( M ) and x < z < y, then z ϵ f ( M )
-I have a feeling the intermediate value theorem will play a part in this, but I'm not sure how.
I'm having trouble relating these three points.

Answer & Explanation

Lara Alvarez

Lara Alvarez

Beginner2022-05-11Added 14 answers

Proof 1: If f ( p ) s for any p M, then
f 1 ( ( s , ) ) f 1 ( ( , s ) ) = M
The two sets on the left are open and disjoint. Since M is connected, one of them is M and one of them is . So either
f 1 ( ( s , ) ) = M p M ,   f ( p ) > s
or
f 1 ( ( , s ) ) = M p M ,   f ( p ) < s
Q.E.D

.Proof 2: Suppose it is not true that
( p M ,   f ( p ) > s )  or  ( q M ,   f ( q ) < s )
Since s is not a value of f, we have
( p M ,   f ( p ) < s )  and  ( q M ,   f ( q ) > s )
By the Intermediate Value Theorem, there exists r M with f ( r ) = s. This contradicts the condition that s is not a value of f. Q.E.D.

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