Continuity of f^+ ... Browder Corollary 3.13

  • #1
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Summary:

I need help in order to demonstrate a formal and rigorous proof that given a real function f is continuous that ##f^+## is also continuous ... ...

Main Question or Discussion Point

I am reading Andrew Browder's book: "Mathematical Analysis: An Introduction" ... ...

I am currently reading Chapter 3: Continuous Functions on Intervals and am currently focused on Section 3.1 Limits and Continuity ... ...

I need some help with the proof of Corollary 3.13 ...


Corollary 3.13 reads as follows:


Browder - Corollary 3.13 ... .png



Can someone help me to prove that if ##f## is continuous then ##f^+ = \text{max} (f, 0)## is continuous ...


My thoughts are as follows:

If ##c## belongs to an interval where ##f## is positive then ##f^+## is continuous since ##f## is continuous ... further, if ##c## belongs to an interval where ##f## is negative then ##f^+## is continuous since ##g(x) = 0## is continuous ... but how do we construct a proof for those points where ##f(x)## crosses the ##x##-axis ... ..



Help will be much appreciated ...

Peter
 

Answers and Replies

  • #2
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Hi Peter,

you got the right point where it might be a problem, namely when ##f## changes to ##0##. Now a proof depends on the definition you use for a continuous function. You used "preimages of open sets are open" recently. So take an open set around such a point where ##f## crosses the ##x-##axis and determine its preimage.
 
  • #3
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A more general thing is true. If ##f,g## are continuous functions, then so is ##f\lor g=\max\{f,g\}##.

The easiest way to see this is to note that

$$f\lor g=\frac{1}{2}(f+g+|f-g|)$$
 
  • #4
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My thanks to Math_QED and fresh_42 for their help ...

Following the advice of fresh_42 my proof of the case where ##f## crosses the axis proceeds as follows:


I think it will suffice to prove that ##f^+## is continuous for the case where a point ##c_1 \in \mathbb{R}## is such that for ##x \lt c_1##, ##f(x) = f^+(x)## is positive and for ##x \gt c_1##, ##f^+(x) = 0## ... ... while for some point ##c_2 \gt c_1## we have that ##f^+(x) = 0## for ##x \lt c_2## and ##f(x) = f^+(x)## is positive for ##x \gt c_2## ... ...

... see Figure 1 below ...


Figure 1 - Continuity of f+ ... .png




Now consider an (open) neighbourhood ##V## of ##f^+(c_1)## where...

##V= \{ f^+(x) \ : \ -f^+(a_1) \lt f^+(c_1) \lt f^+(a_1)## for some ##a_1 \in \mathbb{R} \}##

so ...

##V= \{ f^+(x) \ : \ -f^+(a_1) \lt 0 \lt f^+(a_1)## for some ##a_1 \in \mathbb{R} \}## ...


Then ... (see Figure 1) ...

##(f^+)^{ -1 } (V) = \{ a_1 \lt x \lt a_2 \}## which is an open set as required ....

Further crossings of the x-axis by ##f## just lead to further sets of the nature ##\{ a_{ n-1 } \lt x \lt a_n \}## which are also open ... so ...

##(f^+)^{ -1 } (V) = \{ a_1 \lt x \lt a_2 \} \cup \{ a_3 \lt x \lt a_4 \} \cup \ldots \cup \{ a_{n-1} \lt x \lt a_n \} ##

which being a union of open sets is also an open set ...


The proof is similar if ##f## first crosses the x-axis from below ...



Is that correct?

Peter
 
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  • #5
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You have written ##f##, but we are interested in ##f^+##. And the ordering is deliberate. Who guarantees you that ##f(]x_1,x_2[)=]-y_1,y_1[##? What we want to know is, how the preimage of ##(f^{+})^{-1}(]y_1,y_2[)## looks like.

We start with an open interval ##V\subseteq \mathbb{R}##. This is sufficient as any open set is a union of open intervals, and ##(f^{+})^{-1}(\cup_\iota U_\iota) = \cup_\iota (f^{+})^{-1}(U_\iota)## which is open if the individual sets are. Hence ##V=]y_1,y_2[##. If ##0\neq V##, then ##f^+\equiv f## or ##f^+\equiv 0## which are both continuous and we are done.

Thus we are left with ##0\in V = ]y_1,y_2[##. What is ##(f^{+})^{-1}(V)## and how can we find out?
 
  • #6
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You have written ##f##, but we are interested in ##f^+##. And the ordering is deliberate. Who guarantees you that ##f(]x_1,x_2[)=]-y_1,y_1[##? What we want to know is, how the preimage of ##(f^{+})^{-1}(]y_1,y_2[)## looks like.

We start with an open interval ##V\subseteq \mathbb{R}##. This is sufficient as any open set is a union of open intervals, and ##(f^{+})^{-1}(\cup_\iota U_\iota) = \cup_\iota (f^{+})^{-1}(U_\iota)## which is open if the individual sets are. Hence ##V=]y_1,y_2[##. If ##0\neq V##, then ##f^+\equiv f## or ##f^+\equiv 0## which are both continuous and we are done.

Thus we are left with ##0\in V = ]y_1,y_2[##. What is ##(f^{+})^{-1}(V)## and how can we find out?

Oh ... that was a typo ... made a mistake copying my notes without thinking ... have edited it now ...

Is the post still in error ...?

... will now reflect on what you have written ...

Peter
 
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  • #7
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I do not see where you get this very specific form of ##V## from. We have to show that ##(f^+)^{-1}(V)## is open for any open set ##V\subseteq \mathbb{R}##. See my previous post for why we can assume that ##V## is an interval.

Now if ##0\notin V## then ##(f^+)^{-1}(V)## is either empty or ##(f^+)^{-1}(V)=f^{-1}(V)##, which both are open sets. Thus we may assume that ##0\in V## and we have to determine ##(f^+)^{-1}(V)## for this case. Yes, ##V## looks like ##V=(y_1,0)\cup \{0\} \cup (0,y_2)## in this case, but we cannot know what ##f^+## does there, so we cannot write ## y_1=f^+(a_1)\, , \,y_2=f^+(a_2)\,.## Maybe ##f^+(x)=f(x)> 1000## and we have chosen ##y_2=1##. The task is: Show that ##(f^+)^{-1}((y_1,0)\cup \{0\} \cup (0,y_2))## is open for any pair ##y_1<0<y_2\,.##

If we want to consider continuity especially at ##x=c## where ##f## crosses the ##x-##axis, then you should use the analytical ##\varepsilon-\delta## definition of continuity, not the topological. If you go by open sets, then any open set ##V## is possible.

You can still use @Math_QED 's approach which uses the theorem: certain combinations of continuous functions are again continuous.
 
  • #8
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I do not see where you get this very specific form of ##V## from. We have to show that ##(f^+)^{-1}(V)## is open for any open set ##V\subseteq \mathbb{R}##. See my previous post for why we can assume that ##V## is an interval.

Now if ##0\notin V## then ##(f^+)^{-1}(V)## is either empty or ##(f^+)^{-1}(V)=f^{-1}(V)##, which both are open sets. Thus we may assume that ##0\in V## and we have to determine ##(f^+)^{-1}(V)## for this case. Yes, ##V## looks like ##V=(y_1,0)\cup \{0\} \cup (0,y_2)## in this case, but we cannot know what ##f^+## does there, so we cannot write ## y_1=f^+(a_1)\, , \,y_2=f^+(a_2)\,.## Maybe ##f^+(x)=f(x)> 1000## and we have chosen ##y_2=1##. The task is: Show that ##(f^+)^{-1}((y_1,0)\cup \{0\} \cup (0,y_2))## is open for any pair ##y_1<0<y_2\,.##

If we want to consider continuity especially at ##x=c## where ##f## crosses the ##x-##axis, then you should use the analytical ##\varepsilon-\delta## definition of continuity, not the topological. If you go by open sets, then any open set ##V## is possible.

You can still use @Math_QED 's approach which uses the theorem: certain combinations of continuous functions are again continuous.

Hi fresh_42 ...

Thanks for your help ...

You write:

" ... ... I do not see where you get this very specific form of ##V## from. ... ... "

I'll try to explain ...

First, the context of my attempt at a proof was to show that ##f^+## was continuous at a point ##c_1## where ##f## crossed the x-axis from above ... that is the situation where point ##c_1 \in \mathbb{R}## is such that for ##x \lt c_1##, ##f(x) = f^+(x)## is positive and for ##x \geq c_1##, ##f^+(x) = 0## ... but I also considered f to be such that for some point ##c_2 \gt c_1## f crosses the x-axis from below ... ... that is for some point ##c_2 \gt c_1## we have that ##f^+(x) = 0## for ##x \leq c_2## and ##f(x) = f^+(x)## is positive for ##x \gt c_2## ... ...

The aim was to generalise from this specific form of ##f## and ##f^+## ... ...


Now for this proof I am following Andrew Browder's book "Mathematical Analysis: An Introduction" ...

Browder defines a neighbourhood of a point ##a \in \mathbb{R}## as follows:


Browder - Defn 3.3 ... Neighbourhood ... .png



and I also followed Browder Proposition 3.9 (d) winch reads as follows:


Browder - Proposition 3.9 ... .png




So following Definition 3.3 and Proposition 3.9 I formed a neighbourhood ##V## of ##f^+(c_1)## as

##V= \{ f^+(x) \ : \ -f^+(a_1) \lt f^+(c_1) \lt f^+(a_1)## for some ##a_1 \in \mathbb{R} \}##

so ...

##V= \{ f^+(x) \ : \ -f^+(a_1) \lt 0 \lt f^+(a_1)## for some ##a_1 \in \mathbb{R} \}## ...



I hope that makes sense...

Peter
 
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  • #9
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Definition 3.3. defines a neighborhood and the subset topology. I wanted to use definition (d) (or (c)) of proposition 3.9. The crucial point is in both cases: for every neighborhood ##V##. So we start with an arbitrary neighborhood of ##0\in V## since ##f(c)=0##. We may assume ##0\in V=(y_1,y_2)##. We may not assume ##y_i=f^+(a_i)## since this wouldn't be an arbitrary ##V##.

Take your picture, rotate it by ##90°## to the left
1579404745259.png

and calculate ##(f^+)^{-1}(V)## for various possibilities of ##f^+##.

1579405234438.png
 
  • #10
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Definition 3.3. defines a neighborhood and the subset topology. I wanted to use definition (d) (or (c)) of proposition 3.9. The crucial point is in both cases: for every neighborhood ##V##. So we start with an arbitrary neighborhood of ##0\in V## since ##f(c)=0##. We may assume ##0\in V=(y_1,y_2)##. We may not assume ##y_i=f^+(a_i)## since this wouldn't be an arbitrary ##V##.

Take your picture, rotate it by ##90°## to the left
View attachment 255805
and calculate ##(f^+)^{-1}(V)## for various possibilities of ##f^+##.

View attachment 255807


I have started to examine some possibilities for ##f^+## ... calculating ##(f^+)^{-1} (V)## for each ...

Examples are as follows:



Figure 2 - Continuity of f+ ... .png



For ##f^+## as shown in Figure 2 we have:

##V= \{ y \ : \ y_2 \lt y \lt y_1 \}##

so that

##(f^+)^{-1} (V) = \{ x \ : \ x_1 \lt x \lt x_2 \}## ... ... ... ... ... which is an open set ...



For a second example consider ##f^+## as shown in Figure 3 below ... ...


Figure 3 - Continuity of f+ ... .png





For ##f^+## as shown in Figure 3 we have:

##V= \{ y \ : \ y_2 \lt y \lt y_1 \}##

so that

##(f^+)^{-1} (V) = \{ x \ : \ - \infty \lt x \lt x_2 \}## ... ... ... ... ... which is an open set ...



For a third example consider ##f^+## as shown in Figure 4 below ... ...



Figure 4 -  Continuity of f+ ... .png




For ##f^+## as shown in Figure 4 we have:

##V= \{ y \ : \ y_2 \lt y \lt y_1 \}##

so that

##(f^+)^{-1} (V) = \{ x \ : \ - \infty \lt x \lt x_1 \} \cup \{ x \ : \ x_2 \lt x \lt x_3 \}## ... ... ... ... ... which is an open set ...



I have two questions ....

1. Is the above analysis correct as far as it goes ...?

2. If it is correct ... how do I use the analysis to organize and construct a general proof ... ...



Hope that you can help further ...

Peter
 
  • #11
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Yes, this looks ok so far. But listing all possibilities is a bit inconvenient.

We can assume that ##f^+(\mathbb{R})\cap V \neq \emptyset## since otherwise ##(f^+)^{-1}(V)=\emptyset## which is open and we are done. We can also assume that ##f(c)= 0\in V## for otherwise we have again ##(f^+)^{-1}(V)=\emptyset ## or ##(f^+)^{-1}(V)=f^{-1}(V)## which is open by continuity of ##f##.

Our ##V## looks like ##V=(y_1,0) \cup \{0\} \cup (0,y_2)## with ##y_1<0<y_2## so ##(f^+)^{-1}(V)=(f^+)^{-1}((y_1,0))\cup (f^+)^{-1}(\{0\}) \cup (f^+)^{-1}((0,y_2)=\emptyset \cup (f^+)^{-1}([0,y_2))=(f^+)^{-1}([0,y_2))##.

Now we only have to bother, whether ##y_2\in f(\mathbb{R})## or not. So what is ##(f^+)^{-1}([0,y_2))## in these two cases?
a.) ##y_2=f(x_2)## for some ##x_2\in \mathbb{R}##
b.) ##y_2 > f(\mathbb{R})##
 
  • #12
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Yes, this looks ok so far. But listing all possibilities is a bit inconvenient.

We can assume that ##f^+(\mathbb{R})\cap V \neq \emptyset## since otherwise ##(f^+)^{-1}(V)=\emptyset## which is open and we are done. We can also assume that ##f(c)= 0\in V## for otherwise we have again ##(f^+)^{-1}(V)=\emptyset ## or ##(f^+)^{-1}(V)=f^{-1}(V)## which is open by continuity of ##f##.

Our ##V## looks like ##V=(y_1,0) \cup \{0\} \cup (0,y_2)## with ##y_1<0<y_2## so ##(f^+)^{-1}(V)=(f^+)^{-1}((y_1,0))\cup (f^+)^{-1}(\{0\}) \cup (f^+)^{-1}((0,y_2)=\emptyset \cup (f^+)^{-1}([0,y_2))=(f^+)^{-1}([0,y_2))##.

Now we only have to bother, whether ##y_2\in f(\mathbb{R})## or not. So what is ##(f^+)^{-1}([0,y_2))## in these two cases?
a.) ##y_2=f(x_2)## for some ##x_2\in \mathbb{R}##
b.) ##y_2 > f(\mathbb{R})##



Thanks again for your help, fresh_42 ...

You write:

" ... ...
So what is ##(f^+)^{-1}([0,y_2))## in these two cases?
a.) ##y_2=f(x_2)## for some ##x_2\in \mathbb{R}##
b.) ##y_2 > f(\mathbb{R})## ... ... "


Based on what was indicated by the examples I considered ...

... if ##y_2=f(x)## for some ##x \in \mathbb{R}##

... then ...

##(f^+)^{-1} (V) = \{ x \ : \ x_1 \lt x \lt x_2 \} \cup \{ x \ : \ x_3 \lt x \lt x_4 \} \cup \ldots \cup \{ x \ : \ x_{n-1} \lt x \lt x_n \}##

where ##x_1## and/or ##x_n## may have to be replaced by ##- \infty## and ##+ \infty## respectively ... and the number of unions may be countably infinite ...

In all these cases ##(f^+)^{-1} (V)## is an open set ...

... ...

If ##y_2 > f(\mathbb{R})## ...

... then ...

##(f^+)^{-1} (V) = \emptyset## ... ... and ##\emptyset## is an open set ...


Is the above correct?

Peter


*** EDIT ***

Correction to the following:

" ... ...

If ##y_2 > f(\mathbb{R})## ...

... then ...

##(f^+)^{-1} (V) = \emptyset## ... ... and ##\emptyset## is an open set ... ... "


It should read as follows:


If ##y_2 > f(\mathbb{R})## ...

... then ...

##(f^+)^{-1} (V) = \mathbb{R} ## ... ... and ##\mathbb{R}## is an open set ...


Hope that part of my answer is now correct ...

Also suspect that I was in error in not dealing with 0 properly in the expression ##(f^+)^{-1}([0,y_2))##

Peter
 
Last edited:
  • #13
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No, it's not. Don't make it too complicated with the many subintervals. We do not need to and cannot know the details of ##f##. What we need is ##(f^+)^{-1}(V)=(f^+)^{-1}([0,y_2))##.

We also said that we may assume that f touches the ##x-##axis, so we have either ##(f^+)^{-1}([f^+(c),f^+(x)))## or ##(f^+)^{-1}([0,y_2))## where ##y_2> f^+(x)## for all ##x##.

Now forget for a moment the specific situation. Let's assume we have a function ##g: \mathbb{R} \longrightarrow [0,\infty)##. What is ##g^{-1}([g(0),g(x)))## and what is ##g^{-1}([0,C))## where ##\operatorname{im}(g)=g(\mathbb{R})\subseteq [0,C)## since ##C## is bigger than any function value?

What are the pre-images in these cases? Solve it for ##g## and replace ##g## by ##f^+## afterwards.
 
  • #14
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No, it's not. Don't make it too complicated with the many subintervals. We do not need to and cannot know the details of ff. What we need is (f+)−1(V)=(f+)−1([0,y2))(f+)−1(V)=(f+)−1([0,y2)).

We also said that we may assume that f touches the x−x−axis, so we have either (f+)−1([f+(c),f+(x)))(f+)−1([f+(c),f+(x))) or (f+)−1([0,y2))(f+)−1([0,y2)) where y2>f+(x)y2>f+(x) for all xx.

Now forget for a moment the specific situation. Let's assume we have a function g:R⟶[0,∞)g:R⟶[0,∞). What is g−1([g(0),g(x)))g−1([g(0),g(x))) and what is g−1([0,C))g−1([0,C)) where im(g)=g(R)⊆[0,C)im⁡(g)=g(R)⊆[0,C) since CC is bigger than any function value?

What are the pre-images in these cases? Solve it for gg and replace gg by f+f+ afterwards.



I will try to answer your question ...

" ... ... Let's assume we have a function g:R⟶[0,∞). What is g−1([g(0),g(x))) and what is g−1([0,C)) where im(g)=g(R)⊆[0,C) since C is bigger than any function value? ... ... "


Consider g−1([g(0),g(x))) given g as follows:


Figure 5 -  Continuity of f+ .png






We have g−1([g(0),g(x)))=[0,x) ... ...

But not quite sure why we are determining this pre-image ...




Now consider g−1([0,C)) where im(g)=g(R)⊆[0,C) since C is bigger than any function value ... where we are given g as in Figure 6 below



Figure 6 -  Continuity of f+.png




We find g−1([0,C))=R



Can you help further ...

Peter
 
Last edited:
  • #15
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##g^{-1}([g(0),x)) =\{\,u\in \mathbb{R}\,|\,0\leq g(u)< g(x)\,\} = \{\,u\in \mathbb{R}\,|\, g(u)< g(x)\,\}##

We need the definition of ##f^+## faster than I thought, sorry. I have been too concentrated on how to get rid of the boundary ##g(0)## on the left.

What we now have is ##(f^+)^{-1}([(f^+)(0),x)) =\{\,u \in \mathbb{R}\,|\,(f^+)(u)<(f^+)(x)\,\}## and the next step is ##\{\,u \in \mathbb{R}\,|\,(f^+)(u)<(f^+)(x)\,\} = \{\,u \in \mathbb{R}\,|\,f(u)<f(x)\,\}=f^{-1}(\;(-\infty,f(x)\;)## which is open because ##f## is continuous.
 
  • #16
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##g^{-1}([g(0),x)) =\{\,u\in \mathbb{R}\,|\,0\leq g(u)< g(x)\,\} = \{\,u\in \mathbb{R}\,|\, g(u)< g(x)\,\}##

We need the definition of ##f^+## faster than I thought, sorry. I have been too concentrated on how to get rid of the boundary ##g(0)## on the left.

What we now have is ##(f^+)^{-1}([(f^+)(0),x)) =\{\,u \in \mathbb{R}\,|\,(f^+)(u)<(f^+)(x)\,\}## and the next step is ##\{\,u \in \mathbb{R}\,|\,(f^+)(u)<(f^+)(x)\,\} = \{\,u \in \mathbb{R}\,|\,f(u)<f(x)\,\}=f^{-1}(\;(-\infty,f(x)\;)## which is open because ##f## is continuous.

Hi fresh_42 ...

Sorry to be slow ... but what is ##x## in the expression ##g^{-1}([g(0),x))## ... ?

Further ... could you please explain why/how ...

##g^{-1}([g(0),x)) =\{\,u\in \mathbb{R}\,|\,0\leq g(u)< g(x)\,\} = \{\,u\in \mathbb{R}\,|\, g(u)< g(x)\,\}##

Again ... sorry if I am asking the obvious ...

*** EDIT *** ... oh! presumably its just the definition of pre-image ...?

Peter
 
  • #17
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##x## was a very bad choice. It forced me to switch to ##u## as variable name. I made this mistake in post #13.
Better would have been the cases: ##[0,y_2)## with ##(f^+)(c)=0, (f^+)(x_2)=y_2## and ##[0,C)## with ##f^+(x)<C \;\forall\,x\in \mathbb{R}##.

It was meant as a point ##(x,y)=(x_2,y_2=g(x_2))## with some given image point on the right.

Edit: ##x_2## has to be chosen as ##x_2=\sup\{\,x\in \mathbb{R}\,|\,f(x)=y_2\,\}.##
 
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  • #18
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##x## was a very bad choice. It forced me to switch to ##u## as variable name. I made this mistake in post #13.
Better would have been the cases: ##[0,y_2)## with ##(f^+)(c)=0, (f^+)(x_2)=y_2## and ##[0,C)## with ##f^+(x)<C \;\forall\,x\in \mathbb{R}##.

It was meant as a point ##(x,y)=(x_2,y_2=g(x_2))## with some given image point on the right.

OK ... understand ...

Thanks once again for all your help ...

Peter
 
  • #19
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Hi fresh_42 ...

I thought I'd just (for completeness) set out the proof of the continuity of ##f^+## in terms of the variables we were talking about early on ...

Consider Case 1: ##y_2 = f^+ (x_2)## for some ##x_2 \in \mathbb{R}## ... ...

See Figure 7 below ...


Figure 7 -  Continuity of f+ ... .png



We have ##y_2 = f^+ (x_2) = f(x_2)## ... and ... ##f^+(c) = f(c) = 0## ...

We are required to calculate ##(f^+)^{-1} ( [0, y_2) )## ...

We have ...

##(f^+)^{-1} ( [0, y_2) ) = \{ x\ : \ 0 \leq f^+ (x) \lt f^+ (x_2) \}##

... so ... ##(f^+)^{-1} ( [0, y_2) ) = \{ x\ : \ f^+ (x) \lt f^+ (x_2) \}##

... therefore ... ##(f^+)^{-1} ( [0, y_2) ) = \{ x\ : \ f(x) \lt f(x_2) \}##

... and therefore ... ##(f^+)^{-1} ( [0, y_2) ) = f^{-1}( ( - \infty , f(x_2) ) )##

... and ##f^{-1}( ( - \infty , f(x_2) ) )## is an open set since ##f## is continuous ...



Now consider Case 2 where ##y_2 \gt f( \mathbb{R} )## ... see Figure 8 below ...


Figure 8 -  Continuity of f+ .png



In Case 2 we have ##(f^+)^{-1} ( [0, y_2) ) = \mathbb{R}## ...

... and ##\mathbb{R}## is an open set ...


Hopefully the above is correct...

Peter
 

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