Is f(x) an Injective Function? Understanding Proof and Notation

[CaptainAmerica17]

TL;DR

I need help checking my proofs as I self-study math.

I typed this up in Overleaf using MathJax. I'm self-studying so I just want to make sure I'm understanding each concept. For clarification, the notation f^{-1}(x) is referring to the inverse image of the function. I think everything else is pretty straight-forward from how I've written it. Thank you for your help!

 

[member 587159]

You must show that $x_1 \in f^{-1}(f(E)) \iff x_1 \in E$. No need to introduce $x_2$. I see why you did that: to get to the definition of injectivity, but it is confusing.

Notice first that one inclusion always holds, without injectivity. Indeed, $f^{-1}(f(E))$ is the set of all elements in $A$ that get mapped to $f(E)$. Clearly $E$ is contained in this set, so we always have (even without injectivity) $E \subseteq f^{-1}(f(E))$.

For the converse, the injectivity will be crucial.

If $x \in f^{-1}(f(E))$, then you know that $f(x) \in f(E)$. How can you proceed and use the injectivity?

 

[PeroK]

Your proof doesn't make much sense to me. When are two sets equal?

 

[CaptainAmerica17]

You must show that $x_1 \in f^{-1}(f(E)) \iff x_1 \in E$. No need to introduce $x_2$. I see why you did that: to get to the definition of injectivity, but it is confusing.

Notice first that one inclusion always holds, without injectivity. Indeed, $f^{-1}(f(E))$ is the set of all elements in $A$ that get mapped to $f(E)$. Clearly $E$ is contained in this set, so we always have (even without injectivity) $E \subseteq f^{-1}(f(E))$.

For the converse, the injectivity will be crucial.

If $x \in f^{-1}(f(E))$, then you know that $f(x) \in f(E)$. How can you proceed and use the injectivity?

Thank you for the reply! I always forget that to show equality after I have to clearly show that each set is a subset of the other :/
The only next step I can think of is that if f(x) is in f(E), then x is in E, right?

 

[member 587159]

The only next step I can think of is that if f(x) is in f(E), then x is in E, right?

Why? Is this true without injectivity?

 

[CaptainAmerica17]

Why? Is this true without injectivity?

Well if the function is injective then we know that x maps to at most one y = f(x). So I don't think this could be true without inectivity.

 

[member 587159]

Well if the function is injective then we know that x maps to at most one y = f(x). So I don't think this could be true without inectivity.

You get the idea, but at this point it is important to write down everything in huge detail!

What is the definition of $f(E)$? Thus what does $x \in f(E)$ mean?

 

[CaptainAmerica17]

The definition of f(E) is {f(x) : x is in E}.

 

[member 587159]

The definition of f(E) is {f(x) : x is in E}.

Okay, now answer the second question. What does $x\in f(E)$ mean?

 

[CaptainAmerica17]

For x to be an element of f(E) it would have to be equal to some f(x) in f(E), right? Just based on the definition. I'm sorry if I'm being slow.

 

[member 587159]

For x to be an element of f(E) it would have to be equal to some f(x) in f(E), right? Just based on the definition. I'm sorry if I'm being slow.

Yes, so we left of at $f(x)\in f(E)$, so there is an $e\in E$ (do not call this new variable $x$, this name is already used!) with $f(x)=f(e)$.

Can you conclude?

 

[CaptainAmerica17]

Yes, so we left of at $f(x)\in f(E)$, so there is an $e\in E$ (do not call this new variable $x$, this name is already used!) with $f(x)=f(e)$.

Can you conclude?

Well if f(x) = f(e) then x = e.

 

[member 587159]

Well if f(x) = f(e) then x = e.

Yes, so is $x \in E$ true?

 

[CaptainAmerica17]

Yes!

 

[member 587159]

Yes!

Congrats, now your proof works!

 

[CaptainAmerica17]

Congrats, now your proof works!

Wow, I really overcomplicated things XD

 

[member 587159]

Wow, I really overcomplicated things XD

Well, let's check if you learned from it! Show the following statement (if you want to/have the time):

If $f$ is surjective, then $f(f^{-1}(E))=E$.

 

[PeroK]

Congrats, now your proof works!

Are you sure?

 

[member 587159]

Are you sure?

I meant that the OP has now a working proof (not the one he wrote down originally):

If $x \in f^{-1}(f(E))$, then $f(x) \in f(E)$, so there is $e\in E$ with $f(x) = f(e)$. By injectivity, $x=e\in E$. This shows $f^{-1}(f(E)) \subseteq E$, the other inclusion is trivial.

 

[PeroK]

Wow, I really overcomplicated things XD

An alternative method is to show

1) first that for any set $E$ and any function $f$ we have $E \subseteq f^{-1}(f(E))$

2) If $E \subset f^{-1}(f(E))$, then $f$ is not one-to-one.

 

[member 587159]

An alternative method is to show

1) first that for any set $E$ and any function $f$ we have $E \subseteq f^{-1}(f(E))$

2) If $E \subset f^{-1}(f(E))$, then $f$ is not one-to-one.

I prefer the notation $\subsetneq$ instead of $\subset$, but yes, that is a good approach too.

 

[PeroK]

I meant that the OP has now a working proof (not the one he wrote down originally):

If $x \in f^{-1}(f(E))$, then $f(x) \in f(E)$, so there is $e\in E$ with $f(x) = f(e)$. By injectivity, $x=e\in E$. This shows $f^{-1}(f(E)) \subseteq E$, the other inclusion is trivial.

I wasn't disputing that you had a proof!

 

[CaptainAmerica17]

Well, let's check if you learned from it! Show the following statement (if you want to/have the time):

If $f$ is surjective, then $f(f^{-1}(E))=E$.

E is a subset of A or B?

 

[member 587159]

E is a subset of A or B?

Of $B$, or $f^{-1}(E)$ wouldn't make sense.

 

[CaptainAmerica17]

Of $B$, or $f^{-1}(E)$ wouldn't make sense.

Ah, ok. I was drawing out a picture and it wasn't making sense.

 

[CaptainAmerica17]

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 $f(x) = y$.

 

[PeroK]

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:
$$ A = B \ \text{iff} \ (A \subseteq B \ \text{and} \ B \subseteq A)$$

 

[CaptainAmerica17]

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, $f(f^{-1}(E)) \subseteq E$.

 

[PeroK]

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 $f(x) = y$.

 

[CaptainAmerica17]

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, $x =f(x) \in f(f^{-1}(E))$.