Hardly even understand problem statement.

[V0ODO0CH1LD]

Homework Statement

Let f: R → P(R) given by:

f(x) = {y ∈ R; sin y < x}.

If A is such that f(x) = R for every x ∈ A, then:

(i) A = [-1, 1];
(ii) A = (-∞, a] for every a < -1;
(iii) A = [a, ∞) for every a > 1;
(iv) A = (-∞, a] for every a ≤ -1;
(v) A = [a, ∞) for every a ≥ 1;

Homework Equations

I have no idea which equations are relevant to this problem.

The Attempt at a Solution

I guess all I know is that it is not (i), since it would be (i) if f(x) = {y ∈ R; sin y = x}. But I am not even sure of that. Could you guys shed a light on this?

 

[Mark44]

Homework Statement

Let f: R → P(R) given by:

What does P(R) mean?
Also, what does R mean? In the notation above, R usually means the set of real numbers, but below you have f(x) = R. Did you mean f(x) $\in$ R?

f(x) = {y ∈ R; sin y < x}.

If A is such that f(x) = R for every x ∈ A, then:

(i) A = [-1, 1];
(ii) A = (-∞, a] for every a < -1;
(iii) A = [a, ∞) for every a > 1;
(iv) A = (-∞, a] for every a ≤ -1;
(v) A = [a, ∞) for every a ≥ 1;

Homework Equations

I have no idea which equations are relevant to this problem.

The Attempt at a Solution

I guess all I know is that it is not (i), since it would be (i) if f(x) = {y ∈ R; sin y = x}. But I am not even sure of that. Could you guys shed a light on this?

 

[Alesak]

P(R) probably means power set of real numbers, meaning to each real number x it assigns a subset of real numbers. You can imagine that for each number x it assigns a set of numbers, such that sin y < x. You can imagine a plot of sine with horizontal line located at height x, and the f(x) would be the the set where the sine function is lower that this line. Just make a picture. Notation in this example is somewhat confusing.

 

[V0ODO0CH1LD]

What does P(R) mean?
Also, what does R mean? In the notation above, R usually means the set of real numbers, but below you have f(x) = R. Did you mean f(x) $\in$ R?

Sorry R means ℝ. And P(R) means the power set of the real numbers. f(x) = R is the way it is written in the problem statement, I'm not really sure if it means that f(x) $\in$ R.

P(R) probably means power set of real numbers. You can imagine that for each number x it assigns a set of numbers, such that sin y < x. You can imagine a plot of sine with horizontal line located at height x, and the f(x) would be the the set where the sine function is lower that this line. Just make a picture. Notation in this example is somewhat confusing.

It is totally confusing for me! So f: R → P(R) means that f assigns to x a subset of the reals? Isn't that against the definition of a function? Isn't is so that a function can't have more than one output? Or is the one output that whole subset of R? Also, does f(x) = R also means that f(x) belongs to the reals?

Here is a plot of the function:

http://www.wolframalpha.com/input/?i=sin+y+<+x

I still don't understand how f assigns a value (or set of values) for a given x.

 

[Alesak]

It is totally confusing for me! So f: R → P(R) means that f assigns to x a subset of the reals? Isn't that against the definition of a function? Isn't is so that a function can't have more than one output? Or is the one output that whole subset of R? Also, does f(x) = R also means that f(x) belongs to the reals?
Here is a plot of the function:

http://www.wolframalpha.com/input/?i=sin+y+<+x

I still don't understand how f assigns a value (or set of values) for a given x.

What happens when you take power set of some set is that each subset is taken as element of P(S), where S is arbitrary set. For example, power set of S = {1, 2, 3} is {{1}, {2}, {3}, {1, 3}, {2, 3}, {1, 2, 3}}. So function from, say, R to P(x) assigns to each real number a subset of that set, which means element of the set {{1}, {2}, {3}, {1, 3}, {2, 3}, {1, 2, 3}}. You can write f(5) = {2, 3} without any problems.

Don't forget that elements of sets can be sets themselves!

 

[Alesak]

More appropriate mathematica example would be this:

sin x < 0.5

can you show the f(0.5) on that picture?

 

[Ray Vickson]

Sorry R means ℝ. And P(R) means the power set of the real numbers. f(x) = R is the way it is written in the problem statement, I'm not really sure if it means that f(x) $\in$ R.



It is totally confusing for me! So f: R → P(R) means that f assigns to x a subset of the reals? Isn't that against the definition of a function? Isn't is so that a function can't have more than one output? Or is the one output that whole subset of R? Also, does f(x) = R also means that f(x) belongs to the reals?

Here is a plot of the function:

http://www.wolframalpha.com/input/?i=sin+y+<+x

I still don't understand how f assigns a value (or set of values) for a given x.

The function f assigns a set to each x, rather than a number. That is NOT against the general definition of a "function", because the values of a function (i.e., the elements of its range) need not be numbers.

RGV

 

[V0ODO0CH1LD]

More appropriate mathematica example would be this:

sin x < 0.5

can you show the f(0.5) on that picture?

Would that be all the angles whose sin equal less than .5? In which case, f assigns to x all the values whose sin equal less than x?

Therefore; A (in the problem statement) cannot be -1 or less because there are no real values whose sin equal less than -1. Is that right?

But in the answers to the original problem, there are two sets A that would meet those requirements. (iii) and (v).

 

[Alesak]

Would that be all the angles whose sin equal less than .5? In which case, f assigns to x all the values whose sin equal less than x?

Pretty much. More correct way to think about it is that f assigns to each x a subset of R, therefore f(x) is equal to a single element of P(R).

The important concept here is that set can be taken as element of other set.

Therefore; A (in the problem statement) cannot be -1 or less because there are no real values whose sin equal less than -1. Is that right?

But in the answers to the original problem, there are two sets A that would meet those requirements. (iii) and (v).

Exactly. To decide between 3 and 5, notice the strict inequality in definition of f. What set is f(1)?

 

[V0ODO0CH1LD]

Exactly. To decide between 3 and 5, notice the strict inequality in definition of f. What set is f(1)?

All numerical values whose sin is less than 1. But I think both 3 and 5 are satisfactory definitions of A. Because A is such that all x's in it satisfy the condition that f(x) = R. In other words it is not necessary for A to include all values of x such that f(x) = R.

Or is that weird looking equal sign in f(x) = R different from f(x) $\in$ R in which case I can select only one answer. Or did I not get it yet?

 

[LCKurtz]

What happens when you take power set of some set is that each subset is taken as element of P(S), where S is arbitrary set. For example, power set of S = {1, 2, 3} is {{1}, {2}, {3}, {1, 3}, {2, 3}, {1, 2, 3}}.

I'm a bit late to the thread and it doesn't really affect the point you are making, but you are missing two elements of S: {1, 2} and { }. You need 8 elements.:uhh:

 

[Alesak]

Or is that weird looking equal sign in f(x) = R different from f(x) $\in$ R in which case I can select only one answer. Or did I not get it yet?

Correct way to write it is f(x) $\in$ P(R). If you think about it a little bit more, you will surely understand it very clearly. Read article on wikipedia about power sets, if you need.

All numerical values whose sin is less than 1. But I think both 3 and 5 are satisfactory definitions of A. Because A is such that all x's in it satisfy the condition that f(x) = R. In other words it is not necessary for A to include all values of x such that f(x) = R.

You almost got it;)

The answer lies in the expression

f(x) = {y ∈ R; sin y < x}.


If it was f(x) = {y ∈ R; sin y <= x}, both 3 and 5 would be valid.

If you have {y ∈ R; sin y < 1}, does it include pi\2?

 

[Alesak]

I'm a bit late to the thread and it doesn't really affect the point you are making, but you are missing two elements of S: {1, 2} and { }. You need 8 elements.:uhh:

You are right, a mistake.

 

[V0ODO0CH1LD]

Correct way to write it is f(x) $\in$ P(R). If you think about it a little bit more, you will surely understand it very clearly. Read article on wikipedia about power sets, if you need.



You almost got it;)

The answer lies in the expression

f(x) = {y ∈ R; sin y < x}.


If it was f(x) = {y ∈ R; sin y <= x}, both 3 and 5 would be valid.

If you have {y ∈ R; sin y < 1}, does it include pi\2?

Okay, first a couple questions:

Did you mean that f(x) = R is the same as f(x) $\in$ P(R)? Is that because f(x) = R implies that f(x) is equal to the set of real numbers and all the subsets within it? I don't quite get why the notation f(x) = R implies f(x) $\in$ P(R).

Does A = [a, ∞) for every a > 1; mean that A contains all values for x that such that 1 < x < ∞ in a more complicated way?

And finally; do you mean that if I select the A where 1 ≤ x < ∞ that would allow x to equal 1. And since "sin y < x" there would be a value of x (1) that would not include the value pi/2 for y and therefore not include all real values that y can take on?

If that is the case, than what is confusing me is that the problem says that f(x) has to only take real values for all x in A. Not that f(x) must take all real values that sin y can take for every x in A. Or is that wrong because I still don't get the f(x) = R?

 

[Alesak]

Okay, first a couple questions:

Did you mean that f(x) = R is the same as f(x) $\in$ P(R)? Is that because f(x) = R implies that f(x) is equal to the set of real numbers and all the subsets within it? I don't quite get why the notation f(x) = R implies f(x) $\in$ P(R).

It got a bit messy, so let me explain. When you write f(x) $\in$ R, it means f(x) is real number, i.e. the output from the function is a number. But when you write f(x) $\in$ P(R), it means f(x) takes values in P(x), meaning output of f is element of P(R), meaning output of f is a subset of R.

Keep in mind that R is an element of P(R), since R is subset of R. So f(x) = R means merely that f(x) point to specific element of P(R) which in this case happens to be R. We could write for example f(x) = (0,1) or f(x) = {10, 12} for other cases, as all these are subsets of R.

Does A = [a, ∞) for every a > 1; mean that A contains all values for x that such that 1 < x < ∞ in a more complicated way?

And finally; do you mean that if I select the A where 1 ≤ x < ∞ that would allow x to equal 1. And since "sin y < x" there would be a value of x (1) that would not include the value pi/2 for y and therefore not include all real values that y can take on?

If that is the case, than what is confusing me is that the problem says that f(x) has to only take real values for all x in A. Not that f(x) must take all real values that sin y can take for every x in A. Or is that wrong because I still don't get the f(x) = R?

The assignment is needlesly confused with set A. It could have read:


Let f: R → P(R) given by:

f(c) = {x ∈ R; sin x < c}.

Find all c such that f(c) = R.


I have relabeled it a bit since x is usualy horizontal axis. Try to solve this problem. It is really helpful here to make a nice picture, similar to one I posted in mathematica. From this, you will see that c needs to be larger than 1.

Then you will see which A is right.

 

[V0ODO0CH1LD]

AH! f(c) = R specifies that f(c) must equal the set of all real numbers. Which is a valid statement because f is a function such that f:R → P(R) and P(R) includes the set of all real numbers; right?

Now, f(c) = R is a condition on c. So c must be such that all values of f(c) equal the set of all real numbers. And if c = 1 then f(c) could never equal the values of x for which sin x = 1 which are real numbers and therefore f(c) could never equal the set of all real numbers.

So the answer is 3? Also; does A = [a, ∞) for every a > 1 mean A = (a, ∞)?