# Thread: Proof on Openness of a Subset and a Function of This Subset

1. ## 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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2. 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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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

3. 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.

4. 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