What Is the Intersection of Subsets in Real Analysis?

[phillyolly]

Homework Statement

The problem is attached. Please help me out in understanding this problem. This is not a HW question, just for my own understanding...

Homework Equations

The Attempt at a Solution

 

[phillyolly]

So, first, I would like to show that E intersects with F in 0.
Since E= -1>=x>=0 and F E= 0>=x>=1, these two intervals overlap only in 0.

 

[phillyolly]

What is f(E) in this question?

 

[lanedance]

do you have a question about the problem? Even if its not for homeowrk, you should attempt it & get lead through

 

[phillyolly]

Here was I tried to solve. I found that f(E) and f(F) are the same. I don't get that f(E overlap F)=0.

 

[Raskolnikov]

Well you showed that E \cap F = \{ 0 \}. So then we simply have

f(E \cap F) = f(\{ 0 \}) = f(0) = 0.

I don't get where you're getting lost. Anything in specific?
You also already showed that f(E) = f(F) = {y : 0 <= y <= 1}, so that parts good.

All that's left is for you to answer the following: "What would happen if 0 is deleted from the sets E and F?"
What would E intersect F be? What would f(E intersect F) be? Would f(E) still equal f(F)? And probably most importantly: would we still have f(E \cap F) \subset ( f(E) \cap f(F) ) ?

 

[phillyolly]

Answering your questions,
(E intersect F)=N/A,
f(E intersect F)=N/A,
f(E) will still be equal f(F),
And the last question is tricky for me.

 

[annoymage]

you already know <br /> f(E \cap F)=\left\{\right\},\emptyset, empty set<br /> and ( f(E) \cap f(F) ) =\left\{y\inR:0&lt;y\leq1\right\}<br />

so the question are <br /> \emptyset \subset \left\{y\inR:0&lt;y\leq1\right\}<br />??

 

[phillyolly]

No, there is no empty set in 0=<y=<1...
Thank you...

 

[annoymage]

i don't know this is definition or theorem, because i didn't take rigorous set theory yet.

Definition.
Empty set is the set having no element, and it is a subset of every set

the answer to <br /> <br /> \emptyset \subset \left\{y\inR:0&lt;y\leq1\right\}<br /> <br /> is true.

even <br /> <br /> \emptyset \subset \emptyset<br /> <br /> is also true for your information because empty set itself is a set

 

[phillyolly]

This is very helpful for a dummie like me, thank you a lot.

 

[Mark44]

So, first, I would like to show that E intersects with F in 0.
Since E= -1>=x>=0 and F E= 0>=x>=1, these two intervals overlap only in 0.

No, E = {x | -1 <= x <= 0} and F = {x | 0 <= x <= 1}

As you wrote E, it must be true that -1 >= 0, which is not true. For F, you have 0 >= 1, which is also not true.

 

[Raskolnikov]

Mark44, that was a typo on his part. Look at his work in the latest attached thumbnail and his posts since then. He's got it pretty much now I think.

 

[Mark44]

Answering your questions,
(E intersect F)=N/A,
f(E intersect F)=N/A,
f(E) will still be equal f(F),
And the last question is tricky for me.

Your first two answers are incorrect. E \cup F = {0}. This is not the empty set. As Raskolnikov already mentioned, f(E \cup F) = f(0) = 0, which is also not the empty set.

When the problem asks about f(E), it is asking about the interval along the y-axis that the set E is mapped to. IOW, f(E) = {y | y = f(x) for some x in E}.

 

[Mark44]

Mark44, that was a typo on his part. Look at his work in the latest attached thumbnail and his posts since then. He's got it pretty much now I think.

The work in the attached file in post 5 looks pretty good, but post 7, which came later, has some errors, so I'm not so sure the OP has it quite yet.

 

[annoymage]

hmm, mark44, i think
that answer was referred to "What would happen if 0 is deleted from the sets E and F?" and its
E\cap F
;P

 

[Mark44]

I came late to the party, so I missed it that we were removing 0 from the intersection. Sorry.