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

#### Peter

##### Well-known member
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I am reading Houshang H. Sohrab's book: "Basic Real Analysis" (Second Edition).

I am focused on Chapter 4: Topology of [FONT=MathJax_AMS]R[/FONT] 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 $$\displaystyle f^{ -1 } (O') = S \cap O$$ for some open set $$\displaystyle O$$ ... ... "

Can someone please explain why the above quoted statement is true ...

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***EDIT *** ... ... My thoughts on this matter so far ...

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

$$\displaystyle f( S \cap B_\delta ( x_0 ) ) \subseteq B_\epsilon ( f(x_0) ) \subseteq O'$$

Now ... take inverse image under $$\displaystyle f$$ of the above relationship (is this a legitimate move?)

then we have ...

$$\displaystyle S \cap B_\delta ( x_0 ) \subseteq f^{ -1 } ( B_\epsilon ( f(x_0) ) ) \subseteq f^{ -1 } ( O' )$$

So that ... if we put the open set $$\displaystyle B_\delta ( x_0 )$$ equal to $$\displaystyle O''$$ then we get

$$\displaystyle f^{ -1 } ( O' ) \supseteq S \cap O''$$ ...

But now ... how do we find $$\displaystyle O$$ such that

$$\displaystyle f^{ -1 } ( O' ) = S \cap O$$ ...

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Help will be appreciated ...

Peter

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