Question about proof from a guy with a highschool education

[micromass]

Personal learning-method question: Would you guys think it's a good idea if I get started with calculus at the same time I'm learning the naive set theory basics and eventually beyond? Or is that unreasonable to try to learn them simultaneously?

May I ask which book you'll be using for calculus?

 

[reenmachine]

May I ask which book you'll be using for calculus?

Unfortunately Lang's "A first course in calculus'' was too expensive for me at the moment so I bought ''Calculus: An Intuitive and Physical Approach'' by Morris Kline which was the other book you suggested to me in pm.I'll give it a shot and if I don't like it I'll eventually buy Lang's book.It's a gamble but my budget was a little bit tight this week.

My goal is to manage to get through ''Calculus: An Intuitive and Physical Approach'' and then move on to Spivak , which is very expensive but apparently worth it.

 

[WannabeNewton]

You know it's pretty beautiful seeing your progress from page 1 up till now. You are one exceptionally fast learner. Keep it up mate! And remember, topology is the best :D

 

[reenmachine]

You know it's pretty beautiful seeing your progress from page 1 up till now. You are one exceptionally fast learner. Keep it up mate! And remember, topology is the best :D

Thanks , your encouragements means a lot to me!

The biggest reason I'm progressing so smoothly is the quality of the help I'm receiving in the thread.Not just the quality but also the quantity.I can always expect a solid explanation within 12 hours from one of you guys and learning is way easier with some momentum.You guys have been sharp and patient with me as I tend to ask a lot of questions so I'm really grateful for that.Can't thank you guys enough.

Why the topology comment?

 

[micromass]

Thanks , your encouragements means a lot to me!

The biggest reason I'm progressing so smoothly is the quality of the help I'm receiving in the thread.Not just the quality but also the quantity.I can always expect a solid explanation within 12 hours from one of you guys and learning is way easier with some momentum.You guys have been sharp and patient with me as I tend to ask a lot of questions.

Why the topology comment?

Wbn is a bit obsessed by topology. I can't blame him, topology really is the most beautiful part of mathematics. Once you know topology, your definition of beauty changes radically

Now I'm going to get a lot of hate by the people who dislike topology...

 

[WannabeNewton]

Wbn is a bit obsessed by topology. I can't blame him, topology really is the most beautiful part of mathematics. Once you know topology, your definition of beauty changes radically

A bit? Sometimes I dream of null homotopic curves. Also, if I could I would marry Urysohn's lemma :D

 

[pwsnafu]

Thanks , your encouragements means a lot to me!

The biggest reason I'm progressing so smoothly is the quality of the help I'm receiving in the thread.Not just the quality but also the quantity.

I biggest reason is you are asking questions! Too many university students just passively sit in lectures and tutorials, and they never ask anything. There is no such thing as a stupid question.

Why the topology comment?

Topology is very popular with pure mathematicians.

 

[reenmachine]

The only things I know about topology is it deals with shapes , like the difference between a sphere and a torus and of course the poincare conjecture proved by Perelman.

 

[pwsnafu]

The only things I know about topology is it deals with shapes , like the difference between a sphere and a torus and of course the poincare conjecture proved by Perelman.

Topology asks the question "What do continuous functions preserve?" For example f(x) = 2x. It has an inverse g(x) = x/2, which (and this is critical) is also continuous. Now take (say) an open interval. I'll choose (0,1) And now I apply f to it. I get another open interval (0,2). Similarly, I can take (0,2), apply g and obtain (0,1). Hence "Openness" is preserved.

There are other concepts of course.

 

[WannabeNewton]

The only things I know about topology is it deals with shapes , like the difference between a sphere and a torus and of course the poincare conjecture proved by Perelman.

Just to tie things in with the theme of this thread, there is this topology problem involving proving properties of this topological space called the long line which is constructed in a very non-intuitive but elegant way using the Well Ordering Theorem. Anyways, so I still haven't been able to do it and micromass always picks on me for it

 

[Fredrik]

The biggest reason I'm progressing so smoothly is the quality of the help I'm receiving in the thread.Not just the quality but also the quantity.I can always expect a solid explanation within 12 hours from one of you guys and learning is way easier with some momentum.You guys have been sharp and patient with me as I tend to ask a lot of questions so I'm really grateful for that.Can't thank you guys enough.

I think other members here could learn a lot from how you've been handling this. You're getting a lot of quality help because you've been following our recommendations. You have read things we have recommended. You have done exercises that we have recommended. You have answered our questions to you. You have given us feedback on our replies, so that we know what parts you understood and what parts you didn't. Since you're new here, you probably don't realize how rare that is.

A bit? Sometimes I dream of null homotopic curves. Also, if I could I would marry Urysohn's lemma :D

Hahaha...you're funny. I'm not as fanatical about topology as you. It may be because I was naive enough to start reading a book on functional analysis that required topology, before I had studied topology. And then I kept underestimating how much time I would need to spend on topology before I could return to that book. I eventually choose to study other books on functional analysis, even though I did eventually learn some topology.

I have to agree about Urysohn's lemma though. When I had studied the proof, I was in awe of how beautiful it was. Well, the main trick anyway. It was a bit tedious to finish the proof after the main part of it.

The only things I know about topology is it deals with shapes , like the difference between a sphere and a torus and of course the poincare conjecture proved by Perelman.

Mathematicians always mention those things (only those things) when they try to explain what topology is. I find that very strange, because if you study topology, you won't be concerned with any of those things for at least a couple of months. I would say that topology is anything that has anything to do with limits. It's a huge subject that includes the things you mentioned (because the definition of continuity involves limits, and because of some things that are too complicated to explain here).

 

[reenmachine]

I think other members here could learn a lot from how you've been handling this. You're getting a lot of quality help because you've been following our recommendations. You have read things we have recommended. You have done exercises that we have recommended. You have answered our questions to you. You have given us feedback on our replies, so that we know what parts you understood and what parts you didn't. Since you're new here, you probably don't realize how rare that is.

I was unaware it was rare , thought it was fairly common.

Hopefully I can maintain my pace and learn more and more math!

 

[reenmachine]

FINALLY I received my books! Yesss!

Calculus: an intuitive and physical approach by Morris Kline is a brick.Intimidating at first , but I'm not one to back down.That is also without knowing if there's some basic high school concepts I'm unaware of that I would need to understand calculus, but that won't stop me even if that's the case , I'll just pick them up by necessity (it's highly likely that the last high school I've been to had a poor math program).

The book of proof looks very nice and a much easier read , but I won't judge a book by it's cover.

I'm very happy to have them , now I should finish the very last section of the textbook I used during the thread and I'll be ready to start these new adventures.

EDIT: Should I try to do some basic exercises on relations and functions? If so , any suggestions?

 

[Fredrik]

Should I try to do some basic exercises on relations and functions? If so , any suggestions?

There are lots of exercises in the book of proof. The more of them you take the time to work through, the better you will become. You will have to choose how many of them you do, but you should at least do a few from each section with exercises. Do as many as it takes to feel comfortable with the concepts.

One of the most useful results involving relations is that if ~ is an equivalence relation on a set X, then each x in X belongs to exactly one equivalence class, so I would recommend that you make sure that you understand the proof of that. (I haven't checked if the book proves it, or leaves it as an exercise. You should either study the book's proof, or do the exercise).

Some of the most useful stuff involving functions is to be able to determine if a function is injective, surjective, bijective, to understand the terms range and preimage, and prove simple theorems about about those things. For example, $f^{-1}(A\cap B)=f^{-1}(A)\cap f^{-1}(B)$. This is really important because books on analysis and topology never include such details in proofs. You are expected to be able to prove those things for yourself.

 

[reenmachine]

Just a little refresher for fun:1) We will attempt to prove that A∪(B∩Z) = (A∪B)∩(A∪Z).

Let x ∈ A∪(B∩Z) be arbitrary.Since x ∈ A∪(B∩Z) , it implies that x ∈ A.Since x ∈ A , it implies that x ∈ (A∪B) , x ∈ (A∪Z) implying that x ∈ (A∪B)∩(A∪Z).

Now let x ∈ (A∪B)∩(A∪Z) be arbitrary.It implies that x ∈ (A∪B) and that x ∈ (A∪Z) , which in turns implies that x ∈ A.Since x ∈ A , it implies that x ∈ A∪(B∩Z) proving that A∪(B∩Z) = (A∪B)∩(A∪Z).2) We will attempt to prove that A–B = A∩B'.

Let x ∈ A and x ∉ B be arbitrary (x ∈ A-B).Since x ∉ B , it implies that x ∈ B' and since x ∈ A , it implies that x ∈ A∩B'.

Now let x ∈ A∩B' be arbitrary.This implies that x ∈ A and that x ∈ B' , which implies that x ∉ B.Since x ∈ A and x ∉ B , it implies that x ∈ A-B proving that A-B = A∩B'.

 

[reenmachine]

(just using this post to play around a little bit)

Let N be the set of all natural numbers.Let x ∈ N be arbitrary.

∀ x ∈ N , we define f(x) = x^3.

If x=2 , then f(2) = 8
If x=3 , then f(3) = 27
If x=4 , then f(4) = 64

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

∀ x and y ∈ N , we define f(x) = (f(y) = y-5) + 3x

If x=2 and y=3 , then f(2) = 4
If x=13 and y=4 , then f(13) = 38

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

Let set B = {1,2}

∀ x ∈ N , we define f(x) = x + B

If x=5 , the f(5) = 8

Here my problem is how can I express that x would add all elements of B? (x+1+2 in this example)

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

If A = {1,2} , then p(A) = {∅,{1},{2},{1,2}}

If A = {1,2} and B = {4,5} , then A × B = {(1,4),(1,5),(2,4),(2,5)}

If R = {(1,5),(2,5)} , then R' is {(1,4),(2,4)}.The domain of the relation is (1,2).The codomain is (4,5) and the range is (5)? Still having a little bit of trouble here with codomain vs range for some reasons.

R^-1 = {(5,1),(5,2)}

 

[CompuChip]

∀ x and y ∈ N , we define f(x) = (f(y) = y-5) + 3x

If x=2 and y=3 , then f(2) = 4
If x=13 and y=4 , then f(13) = 38

That doesn't make sense. You can define f(x, y) = y - 5 + 3x.
But you can't define f to be two different things (y - 5 or 3x, depending on what you call your variable). Also, I suggest including the y between the brackets to indicate that the function value also depends on y. In terms of relations, you could say this is a subset of N x N x N.

Let set B = {1,2}

∀ x ∈ N , we define f(x) = x + B

If x=5 , the f(5) = 8

Here my problem is how can I express that x would add all elements of B? (x+1+2 in this example)

If B is fixed, you could write f(x) = x + 3 :P
Or you could use sum notation:
f(x) = x + \sum_{b \in B} b

If you want to define such a function for all possible subsets of N, you can include that in your notation. A common way of putting it is f_B(x), to indicate that f depends on B, but not really as a variable but more as an arbitrary - but fixed - value.

Still having a little bit of trouble here with codomain vs range for some reasons.

Yes, the codomain is what you specify in the definition, and guarantees that the function will take values in that set. However, it is not required to cover all the values in the set. In your example, 4 has no pre-image. The range is the subset of the codomain that is actually "used", i.e. if f: A \to B is a function then A is its domain, B is the codomain and the range is
\{ y \in B \mid \exists x \in A : f(x) = y \}
which may or may not equal B (i.e. it's a subset of B).

 

[Fredrik]

This seems to be a good time to point out a pecularity with the definition of a function from X into Y as a special kind of subset of X×Y. Suppose that f is a function from X into Y according to the definition you have studied, and that Y is a proper subset of some set Z. Then f is also a subset of X×Z, and it satisfies both conditions in the definition of a function from X into Z. So Z has an equal right to be called the codomain of f.

With this definition of "function", the range is simply the smallest set that can be called the codomain of f.

Note however that there's an alternative (and not equivalent) definition of "function" that ensures that each function has exactly one codomain. Check out my post #122 again. Definition 1 is equivalent to the definition you have studied. Definition 2 is the alternative I'm talking about. These definitions are two different ways to turn the idea of "a rule that associates exactly one element of Y with each element of X" into a precise mathematical concept.

It rarely matters which definition of function we're using. It only affects the precise choice of words we should use in certain statements. For example, I said earlier that a function $f:X\to Y$ is said to be surjective if f(X)=Y. But if we're using definition 1, every function from X into Y is also a function from X into Z, where Z is any set such that $Y\subseteq Z$. So the statement "f is surjective" is ambiguous. We can't tell if it means f(X)=Y or f(X)=Z. This is not a problem if we use definition 2, which ensures that every function has a unique codomain. If we use definition 1, we should (ideally) use the phrase "f is surjective onto Y" instead of "f is surjective".

 

[reenmachine]

Damn , wrote a long response but the update deleted it one way or another

 

[reenmachine]

That doesn't make sense. You can define f(x, y) = y - 5 + 3x.
But you can't define f to be two different things (y - 5 or 3x, depending on what you call your variable). Also, I suggest including the y between the brackets to indicate that the function value also depends on y. In terms of relations, you could say this is a subset of N x N x N.

So f has to be the exact same function as long as you operate in the same ''universe''?

My point was that the 2nd f was just a second different function that was part of another function's mechanism.

If B is fixed, you could write f(x) = x + 3 :P
Or you could use sum notation:
f(x) = x + \sum_{b \in B} b

Very nice! Thank you! Should be useful if you don't know how many elements are in B.

If you want to define such a function for all possible subsets of N, you can include that in your notation. A common way of putting it is f_B(x), to indicate that f depends on B, but not really as a variable but more as an arbitrary - but fixed - value.

Not sure I understand the logic behind this.

Yes, the codomain is what you specify in the definition, and guarantees that the function will take values in that set. However, it is not required to cover all the values in the set. In your example, 4 has no pre-image. The range is the subset of the codomain that is actually "used", i.e. if f: A \to B is a function then A is its domain, B is the codomain and the range is
\{ y \in B \mid \exists x \in A : f(x) = y \}
which may or may not equal B (i.e. it's a subset of B).

Just to get used to what is a new symbol and concept for me , can you explain more in depth what is the role of ∃ here? I know it means ''there exist'' , but I would like to understand his role better in this specific exemple.

 

[reenmachine]

This seems to be a good time to point out a pecularity with the definition of a function from X into Y as a special kind of subset of X×Y. Suppose that f is a function from X into Y according to the definition you have studied, and that Y is a proper subset of some set Z. Then f is also a subset of X×Z, and it satisfies both conditions in the definition of a function from X into Z. So Z has an equal right to be called the codomain of f.

Great! Does it means that set W , if Z ⊆ W , also has an equal right to be called the codomain of f?

 

[Fredrik]

Great! Does it means that set W , if Z ⊆ W , also has an equal right to be called the codomain of f?

Yes, the set Z in my statement was an arbitrary set that had Y as a subset, so what I said about Z also applies to W.

An example may make the definition $f(A)=\{y\in B\,|\,\exists x\in A~~ f(x)=y\}$ easier to understand. Define $f:\mathbb R\to\mathbb R$ by $f(x)=x^2$ for all $x\in\mathbb R$. The range $f(\mathbb R)$ is the set of all y in the codomain such that there's an x in the domain such that y=f(x)=x². In other words, $f(\mathbb R)$ is the set of all real numbers that are the square of a real number. This is the set of all non-negative real numbers. The negative ones are not included, because for example -1 doesn't have the property that there's an x in the domain such that $x^2=-1$.

I prefer to write the definition as $f(A)=\{f(x)|x\in A\}$. It's just easier to remember this way. But this right-hand side should be viewed as an abbreviated notation for the right-hand side in the more complicated definition. Actually that too, should be viewed as an abbreviation for an even more complicated expression. For example, the statement $\exists x\in A~~f(x)=y$ is an abbreviation for $$\exists x~\left(x\in A~\land~\{\{x\},\{x,y\}\}\in f\right).$$ The $\land$ should be read as "and".

 

[Fredrik]

1) We will attempt to prove that A∪(B∩Z) = (A∪B)∩(A∪Z).

Let x ∈ A∪(B∩Z) be arbitrary.Since x ∈ A∪(B∩Z) , it implies that x ∈ A.Since x ∈ A , it implies that x ∈ (A∪B) , x ∈ (A∪Z) implying that x ∈ (A∪B)∩(A∪Z).

The first sentence is a good start. The second is weird. It's not clear what it means (what does "it" refer to?), and the conclusion $x\in A$ doesn't follow from the first sentence. x could be a member of B-A. After the first sentence, the proof should continue something like this: Since $x\in A\cup(B\cap Z)$, we have $x\in A$ or $x\in B\cap Z$. This implies that $x\in A$ or $x\in B$ and $x\in Z$. This implies that $x\in A\cup B$ and $x\in A\cup Z$. This implies that $x\in (A\cup B)\cap(A\cup Z)$.

The second part of the proof has similar problems. In particular, x may not be a member of A.

2) We will attempt to prove that A–B = A∩B'.

Let x ∈ A and x ∉ B be arbitrary (x ∈ A-B).Since x ∉ B , it implies that x ∈ B' and since x ∈ A , it implies that x ∈ A∩B'.

Now let x ∈ A∩B' be arbitrary.This implies that x ∈ A and that x ∈ B' , which implies that x ∉ B.Since x ∈ A and x ∉ B , it implies that x ∈ A-B proving that A-B = A∩B'.

The statement

Since x ∉ B , it implies that x ∈ B'​

is weird. What you meant to say was presumably

Since $x\notin B$, we have $x\in B'$​

or equivalently

$x\notin B$ implies that $x\in B'$.​

There are many ways to say it. Your way of saying it is weird, because you're making me wonder what "it" refers to. Also for any two statements P and Q, the sentences "Since P, Q" and "P implies Q" mean the same thing. So you shouldn't include both the words "since" and "implies". One of them is sufficient.

My proof of $A-B\subseteq A\cap B'$: Let $x\in A-B$ be arbitrary. We have $x\in A$ and $x\notin B$. This implies that $x\in A$ and $x\in B'$. This implies that $x\in A\cap B'$.

 

[reenmachine]

The first sentence is a good start. The second is weird. It's not clear what it means (what does "it" refer to?), and the conclusion $x\in A$ doesn't follow from the first sentence. x could be a member of B-A. After the first sentence, the proof should continue something like this: Since $x\in A\cup(B\cap Z)$, we have $x\in A$ or $x\in B\cap Z$. This implies that $x\in A$ or $x\in B$ and $x\in Z$. This implies that $x\in A\cup B$ and $x\in A\cup Z$. This implies that $x\in (A\cup B)\cap(A\cup Z)$.

The second part of the proof has similar problems. In particular, x may not be a member of A.

I hate making mistakes like this.Should have concentrated more than I did.

 

[CompuChip]

So f has to be the exact same function as long as you operate in the same ''universe''?

f has to be defined unambiguously. You are of course allowed to define a function g(y) = y - 5, but you shouldn't call it f if f is defined in another way.
Also note that the names of variables are completely arbitrary. E.g. if f(x) = 3x + 1 then you can not only substitute x = 3 to get f(3) = 3 * 3 + 1 = 10, but also x = y to get f(y) = 3y + 1 or x = a₀ to get f(a₀) = 3a₀ + 1. So the function g(y) = y - 5 can also be written as g(x) = x - 5. Therefore you are not allowed to say

f(x) = 3x + 1 and f(y) = y - 5​

because that would give two different definitions to the same function.

My point was that the 2nd f was just a second different function that was part of another function's mechanism.

So you could define g(x) = x - 5 and then write f(x, y) = g(y) - 3x.

Not sure I understand the logic behind this.

In the notation I just made it explicit that, except on the variable x, the function f also depends on some other object, namely the set B. Of course if you want to be very formal, you could go

Define f: \mathbb{R} \times \mathscr{P}(\mathbb{R}) \to \mathbb{R} as f(x, B) = x + \sum_{b \in B} b.​

So f is a function that takes a number (x) and a subset of the reals (B). However, in some applications the subset is "fixed" in the sense that you pick some specific B and then define f as above for that specific B, so it is really only a function of x. Still, there is some dependency on B of course, if I pick a different B then in general f will evaluate to a different number - for example the value of f(0) will change. So instead of writing f(x, B) we often write f(x; B) or f_B(x) to indicate that x is really the variable, but the definition also depends on some (earlier) choice of - in this case - B.
I hope that makes it clearer.

 

[micromass]

f has to be defined unambiguously. You are of course allowed to define a function g(y) = y - 5, but you shouldn't call it f if f is defined in another way.
Also note that the names of variables are completely arbitrary. E.g. if f(x) = 3x + 1 then you can not only substitute x = 3 to get f(3) = 3 * 3 + 1 = 10, but also x = y to get f(y) = 3y + 1 or x = a₀ to get f(a₀) = 3a₀ + 1. So the function g(y) = y - 5 can also be written as g(x) = x - 5. Therefore you are not allowed to say

f(x) = 3x + 1 and f(y) = y - 5​

because that would give two different definitions to the same function.

This is an interesting point and I want to continue a bit on it. What CompuChip is saying is that the functions f(x)=x+5 and f(y)=y+5 are the same thing. In textbooks you will often encounter the phrase that x is a "dummy variable". This means that you can change x in any other symbol (that's not used already) and that the function won't change.

Now, why is this? Well, remember that a function is essentially a set. What set is the function f(x)=x+5. By definition is a subset of $\mathbb{R}\times \mathbb{R}$. Which subset?

f=\{(x,x+5)~\vert~x\in \mathbb{R}\}\subseteq \mathbb{R}\times \mathbb{R}

When we claim that the variable x and y can be interchanged, then we actually say that we have the following equality of sets:

\{(x,x+5)~\vert~x\in \mathbb{R}\} = \{(y,y+5)~\vert~y\in \mathbb{R}\}

This equality can be checked directly. Indeed, take an element in the left hand side and show it is in the right hand side and vice versa. For example, take an arbitrary element of the left-hand side, this has the form $(x,x+5)$ for a certain $x\in \mathbb{R}$. Define $y=x$, then $y\in \mathbb{R}$. Furthermore, we have that $(y,y+5)$ is in the right-hand side (by definition) and that $(x,x+5) = (y,y+5)$. So $(x,x+5)$ is in the right-hand side too.
Showing that an element on the right-hand side is in the left-hand side is analogous.

Now, the above is a bit imprecise. Take two functions $f:A\rightarrow B$ and $g:C\rightarrow D$ are the same thing if
\{(x,f(x))~\vert~x\in A\}= \{(x,g(x))~\vert~x\in C\}
This is nice and it works suitably well. But usually we also demand that $A=C$ and that $B=D$. This has a few consequences that can look weird at first. For example, take $f:\mathbb{R}\rightarrow \mathbb{R}:x\rightarrow x^2$ and $g:\mathbb{R}\rightarrow \mathbb{R}^+ = \{x\in \mathbb{R}~\vert~x\geq 0\}$.
We can see easily that the following equality holds:
\{(x,f(x))~\vert~x\in \mathbb{R}^2\}= \{(x,g(x))~\vert~x\in \mathbb{R}\}
so the graphs of the function are the same and the functions behave the same way. However, the functions are not the same thing, since the codomains of the functions are different.
I agree that this is very weird when you first encounter this. But it makes the life of the mathematician a lot easier!

 

[micromass]

FINALLY I received my books! Yesss!

Calculus: an intuitive and physical approach by Morris Kline is a brick.Intimidating at first , but I'm not one to back down.That is also without knowing if there's some basic high school concepts I'm unaware of that I would need to understand calculus, but that won't stop me even if that's the case , I'll just pick them up by necessity (it's highly likely that the last high school I've been to had a poor math program).

The book of proof looks very nice and a much easier read , but I won't judge a book by it's cover.

I'm very happy to have them , now I should finish the very last section of the textbook I used during the thread and I'll be ready to start these new adventures.

EDIT: Should I try to do some basic exercises on relations and functions? If so , any suggestions?

I suggest you make different threads for the different books you read. So make another thread specifically for Kline. I think it will be a bit confusing if you're going to use this thread both for calculus and set theory.

 

[Fredrik]

What CompuChip is saying is that the functions f(x)=x+5 and f(y)=y+5 are the same thing.

I will elaborate a bit on this too. When I realized these things, I started to get really annoyed by the fact that my books and teachers referred to f(x) as a "function". It's not a function. $f$ is the function. $x$ is a variable that represents a real number. $f(x)$ is an element of the codomain of f. So if $f:\mathbb R\to\mathbb R$, then $f(x)$ is a number, not a function.

As I studied more math, I discovered that small abuses of terminology and notation like this are pretty common. Some of them are pretty useful, because sometimes it's just too annoying to say everything exactly right. Sentences would get too long and weird. But I don't think any of the other standard abuses of terminology and notation can cause as much confusion as this one, so I still refuse to abuse this terminology and notation 99.9% of the time.

I have noticed that I'm not the only one who's been confused by it. I've seen evidence of this in many posts in this forum.

So I will not (or at least very rarely) say things like "the function f(x)=sin x", "the function x²" or "the derivative of sin x is cos x". Instead of the first of these phrases, I'll say "the function f:ℝ→ℝ defined by f(x)=sin x for all x in ℝ". The most accurate way to rephrase the second is similar to the first. A slightly less accurate option (because it doesn't mention the domain or codomain), is "the function $x\mapsto x^2$". Instead of the third, I'll say "the derivative of sin is cos".

 

[micromass]

I will elaborate a bit on this too. When I realized these things, I started to get really annoyed by the fact that my books and teachers referred to f(x) as a "function". It's not a function. $f$ is the function. $x$ is a variable that represents a real number. $f(x)$ is an element of the codomain of f. So if $f:\mathbb R\to\mathbb R$, then $f(x)$ is a number, not a function.

As I studied more math, I discovered that small abuses of terminology and notation like this are pretty common. Some of them are pretty useful, because sometimes it's just too annoying to say everything exactly right. Sentences would get too long and weird. But I don't think any of the other standard abuses of terminology and notation can cause as much confusion as this one, so I still refuse to abuse this terminology and notation 99.9% of the time.

I have noticed that I'm not the only one who's been confused by it. I've seen evidence of this in many posts in this forum.

So I will not (or at least very rarely) say things like "the function f(x)=sin x", "the function x²" or "the derivative of sin x is cos x". Instead of the first of these phrases, I'll say "the function f:ℝ→ℝ defined by f(x)=sin x for all x in ℝ". The most accurate way to rephrase the second is similar to the first. A slightly less accurate option (because it doesn't mention the domain or codomain), is "the function $x\mapsto x^2$". Instead of the third, I'll say "the derivative of sin is cos".

I absolutely agree with this. I really don't like the f(x)=x notation. I will almost always use the full $f:X\rightarrow Y$ notation.
I guess I used the notation in my post because CompuChip used that notation already. Whenever somebody uses some notation in a thread, I generally try to use the same notation so it's less confusing.

The f(x)=x is not even the worst. We also have notations such as $y=x^2$. Those are truly horrible. I generally also dislike the notation for diffy eq. such as $y^\prime = y$. But those are used so often that I'm forced to use them.

And then there's Leibniz notation $\frac{dy}{dx}$, shudder...

The absolute worst notation I've ever seen is in a book where the author decides that function evaluation should be performed along the other side. So instead of f(x), he always writes (x)f. What was he thinking when he decided to use that?

 

[reenmachine]

Yes, the set Z in my statement was an arbitrary set that had Y as a subset, so what I said about Z also applies to W.

An example may make the definition $f(A)=\{y\in B\,|\,\exists x\in A~~ f(x)=y\}$ easier to understand. Define $f:\mathbb R\to\mathbb R$ by $f(x)=x^2$ for all $x\in\mathbb R$. The range $f(\mathbb R)$ is the set of all y in the codomain such that there's an x in the domain such that y=f(x)=x². In other words, $f(\mathbb R)$ is the set of all real numbers that are the square of a real number. This is the set of all non-negative real numbers. The negative ones are not included, because for example -1 doesn't have the property that there's an x in the domain such that $x^2=-1$.

I prefer to write the definition as $f(A)=\{f(x)|x\in A\}$. It's just easier to remember this way. But this right-hand side should be viewed as an abbreviated notation for the right-hand side in the more complicated definition. Actually that too, should be viewed as an abbreviation for an even more complicated expression. For example, the statement $\exists x\in A~~f(x)=y$ is an abbreviation for $$\exists x~\left(x\in A~\land~\{\{x\},\{x,y\}\}\in f\right).$$ The $\land$ should be read as "and".

Again , in $f(A)=\{y\in B\,|\,\exists x\in A~~ f(x)=y\}$ , why is ''there exist'' necessary?

When you define it as $f(A)=\{f(x)|x\in A\}$ , is it implied that f(x) transforms x into y?