Undergrad Continuity and Open Sets .... Sohrab, Theorem 4.3.4 .... ....

[Math Amateur]

TL;DR

This thread concerns an aspect of the proof of the theorem that states that a function is continuous on an open set if and only if the inverse image under f of every open set is open

I am reading Houshang H. Sohrab's book: "Basic Real Analysis" (Second Edition).

I am focused on Chapter 4: Topology of R and Continuity ... ...

I need help in order to fully understand the proof of Theorem 4.3.4 ... ... Theorem 4.3.4 and its proof read as follows:

In the above proof by Sohrab we read the following:

" ... ... Therefore $f^{ -1 } (O') = S \cap O$ for some open set $O$ ... ... "Can someone please explain why the above quoted statement is true ...

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My thoughts on this matter so far ...

Since $f$ is continuous at $x_0$ we can find $\delta$ such that

$f( S \cap B_\delta ( x_0 ) ) \subseteq B_\epsilon ( f(x_0) ) \subseteq O'$Now ... take inverse image under f of the above relationship (is this a legitimate move?)then we have ... $S \cap B_\delta ( x_0 ) \subseteq f^{ -1 } ( B_\epsilon ( f(x_0) ) ) \subseteq f^{ -1 } ( O' )$So that ... if we put the open set $B_\delta ( x_0 )$ equal to $O''$ then we get$f^{ -1 } ( O' ) \supseteq S \cap O''$ ...But now ... how do we find $O$ such that $f^{ -1 } ( O' ) = S \cap O$ ...?

Help will be appreciated ...

Peter

 

[member 587159]

This could be wrong, so take the following hint with a grain of salt, because I don't have the time now to try it myself.

I think you can define $O$ to be the union of some open balls that the proof constructed.

 

[Math Amateur]

Thanks Math_QED...

Reflecting on your suggestion...

Peter

 

[pasmith]

I think we are dealing with the restriction of f to S, so f^{-1}(O') \subset S.
Now the \delta constructed by the proof actually depends on x_0, so we can write <br /> f^{-1}(O&#039;) \subseteq \bigcup_{x_0 \in f^{-1}(O&#039;)} S \cap B_{\delta(x_0)}(x_0) = S \cap \bigcup_{x_0 \in f^{-1}(O&#039;)} B_{\delta(x_0)}(x_0) since x_0 \in S \cap B_{\delta(x_0)}(x_0) and the equality follows from the fundamental laws of set algebra.

It remains to show inclusion in the opposite direction.