Proof on Openness of a Subset and a Function of This Subset

• Oct 17th 2010, 04:31 PM
scherz0
Proof on Openness of a Subset and a Function of This Subset
Dear all,

I am having trouble with understanding the intuition behind the following theorem and certain elements of the proof to this theorem.

Therefore, could someone kindly answer my questions on the proof to this theorem on continuity, all written below?

Thank you very much!

scherz0

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http://img707.imageshack.us/img707/2936/tm113p18.jpg
• Oct 17th 2010, 09:43 PM
tonio
Quote:

Originally Posted by scherz0
Dear all,

I am having trouble with understanding the intuition behind the following theorem and certain elements of the proof to this theorem.

Therefore, could someone kindly answer my questions on the proof to this theorem on continuity, all written below?

Thank you very much!

scherz0

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http://img707.imageshack.us/img707/2936/tm113p18.jpg

1) This follows from the very definition of S...

2) If $f(x)\in U$ , then by definition, $x\in S$ ...

3) This is the very definition of continuity in the general context of topology: a function is continuous iff

the inverse image of an open (closed) set is open (closed)

Tonio
• Oct 24th 2010, 06:59 PM
scherz0
Quote:

Originally Posted by tonio
1) This follows from the very definition of S...

2) If $f(x)\in U$ , then by definition, $x\in S$ ...

3) This is the very definition of continuity in the general context of topology: a function is continuous iff

the inverse image of an open (closed) set is open (closed)

Tonio

Thank you for your response, Tonio.

May I ask kindly if you could elaborate on 2)?

I understand from the definition of S that $x\in S \Rightarrow f(x)\in U$.

But how does the converse hold true? It is written above that $f(x)\in U \Rightarrow x\in S$, yet $f(x)$ isn't defined in terms of anything else in the theorem or proof.
• Oct 24th 2010, 10:04 PM
tonio
Quote:

Originally Posted by scherz0
Thank you for your response, Tonio.

May I ask kindly if you could elaborate on 2)?

I understand from the definition of S that $x\in S \Rightarrow f(x)\in U$.

But how does the converse hold true? It is written above that $f(x)\in U \Rightarrow x\in S$, yet $f(x)$ isn't defined in terms of anything else in the theorem or proof.

As it's defined: $S:=\{x\in\mathbb{R}^n\,;\,f(x)\in U\}$ , so $f(x)\in U$ for some $x\in\mathbb{R}^n$ means that $x\in S$ ...

There is no direct or converse here: $f(x)\in U\Longleftrightarrow x\in S$ , by definition of $S$

Tonio