Proof about injectivity

  • #26
ok, I've thought about it about. The goal of this proof should be to show that ##y \in f(f^{-1}(E) \iff y \in E##

So to start with ##y \in f(f^{-1}(E)##. ##y = f(x)## for some ##x \in f^{-1}(E)##

Or maybe since it is surjective, it is best to start with ##y \in E## so that we can show that there's at least one x in ##f^{-1}(E)##, such that ##f(x) = y##.
 
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  • #27
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ok, I've thought about it about. The goal of this proof should be to show that ##y \in f(f^{-1}(E) \iff y \in E##

So to start with ##y \in f(f^{-1}(E)##. ##y = f(x)## for some ##x \in f^{-1}(E)##

Or maybe since it is surjective, it is best to start with ##y \in E## so that we can show that there's at least one x in ##f^{-1}(E)##, such that ##f(x) = y##.
I'd still like to see a proof of the original problem! These are tricky, I think. Again, I recommend the following:

1) Show that we always have ##f(f^{-1}(E)) \subseteq E##.

2) Show that if ##f## is surjective then ##E \subseteq f(f^{-1}(E)) ##.

In other words:

1) ##f^{-1}(E)## are all the points that map into ##E##. If you map them, they all end up in E.

2) If ##f## is surfective, then all the points in ##E## get mapped to by something and that something must be in ##f^{-1}(E)##. Hence all the points in ##E## are in ##f(f^{-1}(E))##.

PS one way to define equality of set is:
$$ A = B \ \text{iff} \ (A \subseteq B \ \text{and} \ B \subseteq A)$$
 
  • #28
ok, I've thought about it about. The goal of this proof should be to show that ##y \in f(f^{-1}(E) \iff y \in E##

So to start with ##y \in f(f^{-1}(E)##. ##y = f(x)## for some ##x \in f^{-1}(E)##
I'd still like to see a proof of the original problem! These are tricky, I think. Again, I recommend the following:

1) Show that we always have ##f(f^{-1}(E)) \subseteq E##.

2) Show that if ##f## is surjective then ##E \subseteq f(f^{-1}(E)) ##.

In other words:

1) ##f^{-1}(E)## are all the points that map into ##E##. If you map them, they all end up in E.

2) If ##f## is surfective, then all the points in ##E## get mapped to by something and that something must be in ##f^{-1}(E)##. Hence all the points in ##E## are in ##f(f^{-1}(E))##.

PS one way to define equality of set is:
$$ A = B \ \text{iff} \ (A \subseteq B \ \text{and} \ B \subseteq A)$$
Thank you for this. It took a while to reply because I had to get caught up on some school work.
Here's what I worked out:
If ##y \in E##, then ##y = f(x)## for some ##x \in f^{-1}(E)##. Clearly, ##f^{-1}(E)## is the set of all points that map into ##E##, so ##f(f^{-1}(E))## will give us all ##y## such that ##y=f(x)## for all ##x \in f^{-1}(E)##. That is all ##y## such that ##y \in E##. Thus, ##f(f^{-1}(E)) \subseteq E##.

If ##y \in f(f^{-1}(E))##, then ##y = f(x)## for some ##x \in f^{-1}(E)##. By surjectivity, ##x \in f^{-1}(E)## maps to some ##y \in E## such that ##y=f(x)##. Thus, ##f(f^{-1}(E)) \subseteq E##.
 
  • #29
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Thank you for this. It took a while to reply because I had to get caught up on some school work.
Here's what I worked out:
If ##y \in E##, then ##y = f(x)## for some ##x \in f^{-1}(E)##.
Why?

Clearly, ##f^{-1}(E)## is the set of all points that map into ##E##, so ##f(f^{-1}(E))## will give us all ##y## such that ##y=f(x)## for all ##x \in f^{-1}(E)##. That is all ##y## such that ##y \in E##. Thus, ##f(f^{-1}(E)) \subseteq E##.
I can't follow this at all.

If ##y \in f(f^{-1}(E))##, then ##y = f(x)## for some ##x \in f^{-1}(E)##. By surjectivity, ##x \in f^{-1}(E)## maps to some ##y \in E## such that ##y=f(x)##. Thus, ##f(f^{-1}(E)) \subseteq E##.
This is not right. Surjectivity means given ##y \in E##, there exists ##x \in A## such that ##f(x) = y##.
 
  • #30
ok, I've thought about it about. The goal of this proof should be to show that ##y \in f(f^{-1}(E) \iff y \in E##

So to start with ##y \in f(f^{-1}(E)##. ##y = f(x)## for some ##x \in f^{-1}(E)##
I'd still like to see a proof of the original problem! These are tricky, I think. Again, I recommend the following:

1) Show that we always have ##f(f^{-1}(E)) \subseteq E##.

2) Show that if ##f## is surjective then ##E \subseteq f(f^{-1}(E)) ##.

In other words:

1) ##f^{-1}(E)## are all the points that map into ##E##. If you map them, they all end up in E.

2) If ##f## is surfective, then all the points in ##E## get mapped to by something and that something must be in ##f^{-1}(E)##. Hence all the points in ##E## are in ##f(f^{-1}(E))##.

PS one way to define equality of set is:
$$ A = B \ \text{iff} \ (A \subseteq B \ \text{and} \ B \subseteq A)$$
Thank you for this. It took a while to reply because I had to get caught up on some school work.
Here's what I worked out:
If ##y \in f(f^{-1}(E))##, then ##y = f(x)## for some ##x \in f^{-1}(E)##. So if you have ##x \in f(f^{-1}(E))##, you have
it will clearly map back to the set ##E##. So we have ##f(f^{-1}(E)) \subseteq E##
Why?



I can't follow this at all.



This is not right. Surjectivity means given ##y \in E##, there exists ##x \in A## such that ##f(x) = y##.
Ok, so I was correct in my original thinking. To properly show surjection, I have to start that part of the proof with ##y \in E##.

For the first issue you found, it seems to me that it clearly follows from the definition of inverse images. I'm not properly communicating it I guess :/ I'll go simpler.

If ##y \in f(f^{-1}(E))## then there is some ##x \in f^{-1}(E)## such that ##y = f(x)##. From this ##x \in f^{-1}(E)## implies that ##f(x) \in E## such that ##y = f(x) \in E##.

If ##y \in E##, by surjection there is some ##x \in f^{-1}(E)## such that ##y = f(x)##. So by definition, ##x =f(x) \in f(f^{-1}(E))##.
 
  • #31
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If ##y \in f(f^{-1}(E))## then there is some ##x \in f^{-1}(E)## such that ##y = f(x)##. From this ##x \in f^{-1}(E)## implies that ##f(x) \in E## such that ##y = f(x) \in E##.

If ##y \in E##, by surjection there is some ##x \in f^{-1}(E)## such that ##y = f(x)##. So by definition, ##x =f(x) \in f(f^{-1}(E))##.
This looks good. However, you ought to structure it a bit better. Especially the second part you need to say up front you assume ##f## is surjective.
 
  • #32
Let ##y \in E##. Assume that ##f## is surjective. There is some ##x \in f^{-1}(E)## such that ##y = f(x)##. So by definition, ##x =f(x) \in f(f^{-1}(E))##.

At least I finally got something understandable. I didn't have nearly as much trouble proving things about inverse images themselves (i.e. ##f^{-1}(G \cup H) = f^{-1}(G) \cup f^{-1}(H)##). This forced me to more properly understand what is actually being said by "injection" and "surjection".

As an aside, if you don't mind answering: I'm starting my first semester of college in the fall (for math, of course). This is one of the reasons, besides my own interest, that I've started working on proof-writing and real analysis on my own time. Would a proof like the one I've written above be passable in an actual course? Or do you think it would be docked credit for not being so well-written? The school I'm attending focuses a lot on research, and I would love to be prepared enough to get involved (even in a minimal capacity). I've been kind of nervous recently thinking about it.
 
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  • #33
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  • #34
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Superheroes tend to do that ;).
I think you are the second one to make a superhero joke with this user :P
 
  • #35
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I think you are the second one to make a superhero joke with this user :P
Us non-superheroes tend to do that ;). Thanks for the setup.
 
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  • #36
Us non-superheroes tend to do that ;). Thanks for the setup.
Lol
 
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