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

[Math Amateur]

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:

View attachment 9108

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

---------------------------------------------------------------------------------------------------------------------------

Help will be appreciated ...

Peter

 

[jvicens]

Dear Peter,

Thank you for reaching out for assistance with understanding Theorem 4.3.4 in Sohrab's book. I will do my best to explain the statement and prove its validity.

First, let's define some terms for clarity:
- f: a continuous function at x_0
- O': an open set containing f(x_0)
- S: the set on which f is defined
- B_\delta(x_0): the open ball centered at x_0 with radius \delta
- B_\epsilon(f(x_0)): the open ball centered at f(x_0) with radius \epsilon
- f^{-1}(A): the inverse image of set A under function f

Now, the statement in question is: f^{-1}(O') = S \cap O for some open set O. To prove this, we need to show that f^{-1}(O') is equal to the intersection of S and some open set O.

The key here is to use the definition of continuity. Since f is continuous at x_0, we can find a \delta such that f(S \cap B_\delta(x_0)) \subseteq B_\epsilon(f(x_0)) \subseteq O'. This means that for any point in the set S \cap B_\delta(x_0), the corresponding point in the range of f will be contained in the open ball B_\epsilon(f(x_0)), which is a subset of O'.

Now, let's take the inverse image of this statement under f. This means that for any point in the set f^{-1}(O'), the corresponding point in the domain of f will be contained in the set S \cap B_\delta(x_0). In other words, f^{-1}(O') \subseteq S \cap B_\delta(x_0).

Next, we can define O'' = B_\delta(x_0) as an open set. This means that f^{-1}(O') \subseteq S \cap O''. But we want to show that f^{-1}(O') = S \cap O for some open set O. To do this, we can define O = O'' \cup (S \cap O'). This is an open set because it is the union of two open sets. Now, we can see that f^{-1}(O') \subseteq S \cap O, and since we already know that f^{-1}(