Proving Proper Subset Relationships: A, B and C

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In summary, the sets are infinite because there is an infinite number of solutions to the equation of the circle.
  • #1
bergausstein
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1.are the following sets a finite sets? if yes why? if no why?

a. the set of points on a given line exactly one unit from a given point on that line.
b. the set of points in a given plane that are exactly one unit from a given point in that plane.
i'm confused with the part saying "one unit from a given point on that line and "one unit from a given point in that plane. it seems to me that these phrases give hints that the sets are finite. please correct me if I'm wrong.

2. show that if A is a proper subset of B and $B\subseteq C$ then, A is a proper subset of of C.

thanks!
 
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  • #2
1.) To simplify matter a bit, consider:

a) How many points are 1 unit away from the origin on the $x$-axis?

b) How many points are one unit away from the origin in the $xy$-plane?
 
  • #3
uhmm.. infinitely many points. so it's inifinite right?
 
  • #4
bergausstein said:
uhmm.. infinitely many points. so it's inifinite right?

If you are unsure, plot the points in both cases. What do you find?

Or, consider the following:

The first case is:

\(\displaystyle |x|=1\)

How many solutions?

The second case is:

\(\displaystyle x^2+y^2=1\)

How many solutions?
 
  • #5
a has 2 elements.
b has 4.
and can you also help me with 2. i can say that in words but i couldn't do it in a general manner.
 
  • #6
Yes for part a) there are 2 points: \(\displaystyle (\pm1,0)\), but there are an infinite number of points on a circle, uncountably infinite from what I understand.

For question 2, I recommend using a Venn diagram.
 
  • #7
markfl how did you know that question B is talking about the equation of the circle? $\displaystyle x^2+y^2=1$ I put this equation on the wolframalpha and it says that there are 4 solutions to this equation. (1,0), (-1,0), (0,1), (-1,0). how is it infinite?
 
  • #8
The equation of the circle is $x^2+y^2=1$, only if the fixed point (the circle's center) is the origin. One of the definitions of a circle is the set of all points a given distance from a fixed point, and we can use the distance formula to get this equation. For part b) we are not restricted to the axes as we are for the first part, where we are considering only a one-dimensional line.
 
  • #9
MarkFL said:
If you are unsure, plot the points in both cases. What do you find?

Or, consider the following:

The first case is:

\(\displaystyle |x|=1\)

How many solutions?

The second case is:

\(\displaystyle x^2+y^2=1\)How many solutions?
Yes.but how do we prove that the points are infinite??
 
  • #10
We can map the points on the circle to a line segment of length $2\pi r$. According to Cantor, this is equinumerous with $\mathbb{R}$.
 
  • #11
bergausstein said:
I put this equation on the wolframalpha and it says that there are 4 solutions to this equation. (1,0), (-1,0), (0,1), (-1,0). how is it infinite?
Those are only the x and y-intecepts of the function. One way to think about it is that it has two variables but only one equation. As you probably learned in middle school algebra, that means the solution set (x,y) of $x^2+y^2=1$ is infinite.
 
  • #12
I will prove that there are at LEAST as many points (x,y) that satisfy:

\(\displaystyle x^2 + y^2 = 1\)

as there are points in the real interval [0,1].

To do this, I will create an injective function \(\displaystyle f\) from [0,1] to the set \(\displaystyle S\), where:

\(\displaystyle S = \{(x,y) \in \Bbb R^2: x^2 + y^2 = 1\}\).

The function I have in mind is this one:

\(\displaystyle f(a) = (a,\sqrt{1 - a^2})\)

(convince yourself this is indeed a function).

First, we verify that \(\displaystyle f([0,1]) \subseteq S\):

Let \(\displaystyle a \in [0,1]\). Then:

\(\displaystyle a^2 + (\sqrt{1 - a^2})^2 = a^2 + 1 - a^2 = 1\)

(note that we have to have \(\displaystyle |a| \leq 1\) for this to work).

This shows that \(\displaystyle f(a) \in S\), for ANY \(\displaystyle a \in [0,1]\). So the image of \(\displaystyle f\) indeed lies within \(\displaystyle S\) as claimed.

Now, suppose \(\displaystyle f(a) = f(b)\) for \(\displaystyle a,b \in [0,1]\). This means:

\(\displaystyle (a,\sqrt{1 - a^2}) = (b,\sqrt{1 - b^2})\).

This means we MUST have \(\displaystyle a = b\) (since if BOTH coordinates are equal, surely the first coordinates are also equal). So f is injective (one-to-one).

Thus for every \(\displaystyle a \in [0,1]\), we have a corresponding UNIQUE point \(\displaystyle f(a) \in S\), so \(\displaystyle S\) MUST be infinite, as it contains an infinite subset.
 

FAQ: Proving Proper Subset Relationships: A, B and C

1. What is a proper subset?

A proper subset is a subset of a set that contains strictly fewer elements than the original set. In other words, all elements in a proper subset are also elements in the original set, but the proper subset does not contain all the elements of the original set.

2. How do you prove a proper subset relationship?

To prove that set A is a proper subset of set B, you must show that every element in A is also in B, but there exists at least one element in B that is not in A. This can be done by listing out the elements of both sets and comparing them, or by using set notation and logical statements.

3. Can a set be a proper subset of itself?

No, a set cannot be a proper subset of itself. This is because a proper subset must have fewer elements than the original set, and if the two sets are equal, they have the same number of elements.

4. What is the difference between a proper subset and a subset?

A subset is a set that contains all the elements of another set, while a proper subset is a subset that contains strictly fewer elements than the original set. In other words, a proper subset is a subset that is not equal to the original set.

5. Can a set have multiple proper subsets?

Yes, a set can have multiple proper subsets. This is because there can be multiple subsets that contain fewer elements than the original set. For example, the set {1,2,3} has the proper subsets {1}, {2}, {3}, and {1,2}.

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