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Sets

bergausstein

Active member
Jul 30, 2013
191
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!
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
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?
 

bergausstein

Active member
Jul 30, 2013
191
uhmm.. infinitely many points. so it's inifinite right?
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
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?
 

bergausstein

Active member
Jul 30, 2013
191
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.
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
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.
 

bergausstein

Active member
Jul 30, 2013
191
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?
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
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.
 

solakis

Active member
Dec 9, 2012
322
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??
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
We can map the points on the circle to a line segment of length $2\pi r$. According to Cantor, this is equinumerous with $\mathbb{R}$.
 

eddybob123

Active member
Aug 18, 2013
76
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.
 

Deveno

Well-known member
MHB Math Scholar
Feb 15, 2012
1,967
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.