Proving Openness in Metric Spaces

In summary, we discussed two problems. The first problem involved proving that a given function is a metric by breaking it up into n cases and showing the triangle inequality holds for each case. The second problem involved proving that a given set is an open subset of R^2 by showing that for all points in the set, there exists a ball around that point that stays within the set, using the definitions of open sets and boundary points.
  • #1
muzak
44
0
Hi guys, two problems, first one I understand for the most part, the second one, I do not know how to set up and solve.

Homework Statement


Let X = R[itex]^{n}[/itex] for x = (a[itex]_{1}[/itex],...,a[itex]_{n}[/itex]) and y = (b[itex]_{1}[/itex],...,b[itex]_{n}[/itex]), define
d[itex]_{\infty}[/itex](x,y) = max {|a[itex]_{1}[/itex]-b[itex]_{1}[/itex]|,...,|a[itex]_{n}[/itex]-b[itex]_{n}[/itex]|}. Prove that this is a metric.

Homework Equations


Just the triangle inequality part for this one.

The Attempt at a Solution


I've proven the first 3 properties, not quite sure on the last part.
My attempt was to break it up into n cases by supposing a single
difference as the max in each case but with ellipsis but I wasn't
sure on the exactly how. Here's what I attempted:

The distance between the two points is the largest of the n cases.
Let z = (z[itex]_{1}[/itex],...,z[itex]_{n}[/itex]). Then we have n cases to check.
Case 1: d(x,y) = |a[itex]_{1}[/itex]-b[itex]_{1}[/itex]|
Notice that |a[itex]_{1}[/itex]-z[itex]_{1}|[/itex]|[itex]\leq[/itex]max{|a[itex]_{1}[/itex]-z[itex]_{1}[/itex]|,
(I stopped here because I wasn't sure how to proceed,
should I write out to n cases with the ellipsis or two cases?)
What I was going to show after this was that the triangle inequality
holds for a[itex]_{1}[/itex], a[itex]_{2}[/itex], ..., a[itex]_{n}[/itex].

Homework Statement


Prove that the set S = {(x[itex]_{1}[/itex],y[itex]_{1}[/itex] : x[itex]_{1}[/itex] + y[itex]_{1}[/itex] > 0}
is an open subset of R[itex]^{2}[/itex] in the Euclidean metric.

Homework Equations


Euclidean metric, Schwarz Inequality?, Open Ball.

The Attempt at a Solution


I'm not sure how to proceed with this one at all. Picture-wise,
it'd be the region above the y=-x line, I'm guessing,
and I'm guessing I have to pick some arbitrary point in that
region and calculate a strict inequality to show that this is open.
But I do not know how to proceed at all, I'm looking for some hand-holding
at this point really because I want to understand it step by step.

Thanks for any help!
 
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  • #2
mohaque said:
Hi guys, two problems, first one I understand for the most part, the second one, I do not know how to set up and solve.

Homework Statement


Let X = R[itex]^{n}[/itex] for x = (a[itex]_{1}[/itex],...,a[itex]_{n}[/itex]) and y = (b[itex]_{1}[/itex],...,b[itex]_{n}[/itex]), define
d[itex]_{\infty}[/itex](x,y) = max {|a[itex]_{1}[/itex]-b[itex]_{1}[/itex]|,...,|a[itex]_{n}[/itex]-b[itex]_{n}[/itex]|}. Prove that this is a metric.


Homework Equations


Just the triangle inequality part for this one.


The Attempt at a Solution


I've proven the first 3 properties, not quite sure on the last part.
My attempt was to break it up into n cases by supposing a single
difference as the max in each case but with ellipsis but I wasn't
sure on the exactly how. Here's what I attempted:

The distance between the two points is the largest of the n cases.
Let z = (z[itex]_{1}[/itex],...,z[itex]_{n}[/itex]). Then we have n cases to check.
Case 1: d(x,y) = |a[itex]_{1}[/itex]-b[itex]_{1}[/itex]|
Notice that |a[itex]_{1}[/itex]-z[itex]_{1}|[/itex]|[itex]\leq[/itex]max{|a[itex]_{1}[/itex]-z[itex]_{1}[/itex]|,
(I stopped here because I wasn't sure how to proceed,
should I write out to n cases with the ellipsis or two cases?)
What I was going to show after this was that the triangle inequality
holds for a[itex]_{1}[/itex], a[itex]_{2}[/itex], ..., a[itex]_{n}[/itex].

Let's prove this one like this. Can you first show that

[tex]|a_k-b_k|\leq \max \{|a_1-z_1|,...,|a_n-z_n|\}+\max \{|z_1-b_1|,...,|z_n-b_n|\}[/tex]

??

Homework Statement


Prove that the set S = {(x[itex]_{1}[/itex],y[itex]_{1}[/itex] : x[itex]_{1}[/itex] + y[itex]_{1}[/itex] > 0}
is an open subset of R[itex]^{2}[/itex] in the Euclidean metric.


Homework Equations


Euclidean metric, Schwarz Inequality?, Open Ball.


The Attempt at a Solution


I'm not sure how to proceed with this one at all. Picture-wise,
it'd be the region above the y=-x line, I'm guessing,
and I'm guessing I have to pick some arbitrary point in that
region and calculate a strict inequality to show that this is open.
But I do not know how to proceed at all, I'm looking for some hand-holding
at this point really because I want to understand it step by step.

Thanks for any help!

OK, what does open mean for you?? You need to show for al (x,y) in the set that there exists a ball around(x,y) that stays in the set. How would you choose that ball??
Does it makes sense what I'm saying?
 
  • #3
micromass said:
Let's prove this one like this. Can you first show that

[tex]|a_k-b_k|\leq \max \{|a_1-z_1|,...,|a_n-z_n|\}+\max \{|z_1-b_1|,...,|z_n-b_n|\}[/tex]

??OK, what does open mean for you?? You need to show for all (x,y) in the set that there exists a ball around(x,y) that stays in the set. How would you choose that ball??
Does it makes sense what I'm saying?

[itex]|a_k-z_k+z_k-b_k|[/itex][itex]\leq |a_k-z_k|+|z_k-b_k|\leq[/itex] max[itex]{|a_1-z_1|,...,|a_n-z_n|}[/itex]+max{|[itex]z_1-b_1|,...,|z_n-b_n[/itex]|}?I guess for me right now, open for is something without its boundary points. I would have to pick an r such that for some [itex](x_2, y_2), (x_1, y_1) then (x_2-x_1)^{2} + (y_2-y_1)^{2} \leq r^{2}[/itex]?
 
Last edited:
  • #4
mohaque said:
[itex]|a_k-z_k+z_k-b_k|[/itex][itex]\leq |a_k-z_k|+|z_k-b_k|\leq[/itex] max[itex]{|a_1-z_1|,...,|a_n-z_n|}[/itex]+max{|[itex]z_1-b_1|,...,|z_n-b_n[/itex]|}?

OK, and now that the maximum of the left-hand side.

I guess for me right now, open for is something without its boundary points. I would have to pick an r such that for some [itex](x_2, y_2), (x_1, y_1) then (x_2-x_1)^{2} + (y_2-y_1)^{2} \leq r^{2}[/itex]?

What is your definition of open? Saying that it's something without boundary isn't good enough unless you specify what a boundary point is.
 
  • #5
The max's are the distances, gotcha, thanks.

In this problem, I'd say the boundary is the y = -x line.
 
  • #6
mohaque said:
The max's are the distances, gotcha, thanks.

In this problem, I'd say the boundary is the y = -x line.

Yes, but what is the definition of a boundary??
 
  • #7
Set of points that encloses a space, endpoints? I don't know what you mean.
 
  • #8
Please, look up in your course the definition of "open set" and such things. We can't really do much unless we have some definitions to work with...
 
  • #9
It's just the open ball centered at x

B(x,r) = { y in X: d(x,y) strictly less than r}
 
  • #10
mohaque said:
It's just the open ball centered at x

B(x,r) = { y in X: d(x,y) strictly less than r}

What is just the open ball??

Could you please quote the exact definition of an open set?? Yes, it has something to do with open balls.
 
  • #11
Just quoting the definition from my notes:

Given a metric space (X,d) a subset S[itex]\subseteq[/itex]X is called open provided that whenever x[itex]\epsilon[/itex]S, then there is an r>0 such that B(x;r)[itex]\subseteq[/itex]S.
 
  • #12
mohaque said:
Just quoting the definition from my notes:

Given a metric space (X,d) a subset S[itex]\subseteq[/itex]X is called open provided that whenever x[itex]\epsilon[/itex]S, then there is an r>0 such that B(x;r)[itex]\subseteq[/itex]S.

Exactly!

So for our [itex]S=\{(x,y)~\vert~x+y>0\}[/itex], we need to find for each [itex](x,y)\in S[/itex], a r such that [itex]B(x,r)\subseteq S[/itex].

So, let's pick an arbitrary (x,y) in S. What do you think we should take as our r?
 
  • #13
By the way, did you see the open set characterization of continuity yet?? That might come in very handy!
 
  • #14
Something greater than 0 but less than x+y? I don't know how we go about picking the r.

I'm just starting out in real analysis, haven't come across it yet, we've only gotten past closed sets which was a bit easier to digest.
 
  • #15
mohaque said:
Something greater than 0 but less than x+y? I don't know how we go about picking the r.

I'm just starting out in real analysis, haven't come across it yet, we've only gotten past closed sets which was a bit easier to digest.

OK, take an arbitrary point in S. What is the distance between that point and the line y=-x??
 
  • #16
[itex]\sqrt{(y_{o}-y)^{2}+(x_{0}-x})^2[/itex], since y=-x, do I make that substitution into the formula?
 
  • #17
mohaque said:
[itex]\sqrt{(y_{o}-y)^{2}+(x_{0}-x})^2[/itex], since y=-x, do I make that substitution into the formula?

What is [itex](x_0,y_0)[/itex] and (x,y)??
 
  • #18
(x,y),(xo,yo) in S
 
  • #19
mohaque said:
(x,y),(xo,yo) in S

But why would you do that?? Why would you pick both of them in S??

I want to find the least distance between a point (x,y) in S and the line y=-x. This is just a calculus/geometry question... How would you solve that??
 
  • #20
I was thinking we had to use the eclidean distance formula for some reason. |x+y|/sqrt(2)
 
  • #21
mohaque said:
I was thinking we had to use the eclidean distance formula for some reason. |x+y|/sqrt(2)

Indeed, the distance between a point [itex](x,y)\in S[/itex] and the line y=-x is exactly

[tex]\frac{|x+y|}{\sqrt{2}}[/tex]

Now, if we choose

[tex]r=\frac{|x+y|}{\sqrt{2}}[/tex]

isn't it true that the ball B((x,y),r) stays inside of S?
 
  • #22
Sorry for the departure.

Yes, because 0<[itex]\frac{|x+y|}{\sqrt{2}}[/itex]<y1+y2. Thus B((x,y),r)[itex]\subseteq[/itex]S proving the set is an open subset in R[itex]^{2}[/itex].

I see. So when the question asks take the Euclidean metric, I shouldn't use the euclidean metric we learned in class but just find the distance from previous knowledge?

So if I had to prove the set S={(y1,y2): y1>0} is an open subset of R[itex]^{2}[/itex].
Then all I'd have to do is let (x,y)[itex]\epsilon[/itex]S and pick r = |x|, then B((x,y),r)[itex]\subseteq[/itex]S making it an open subset?
 
  • #23
mohaque said:
I see. So when the question asks take the Euclidean metric, I shouldn't use the euclidean metric we learned in class but just find the distance from previous knowledge?

You did use the Euclidean metric here. To find out the distance between the line and the poitn, you actually needed to use the Euclidean metric/

So if I had to prove the set S={(y1,y2): y1>0} is an open subset of R[itex]^{2}[/itex].
Then all I'd have to do is let (x,y)[itex]\epsilon[/itex]S and pick r = |x|, then B((x,y),r)[itex]\subseteq[/itex]S making it an open subset?

The radius depends on the open set in question. Right now [itex]|x+y|/\sqrt{2}[/itex] works, but such a trick doesn't always work...
 
  • #24
Did I mess up the second one then?

I have a question on notation. My professor uses d[itex]_{\infty}[/itex](x,y) = max{|stuff|, ...}. So when the question says prove that the disk x[itex]^{2}[/itex]+y[itex]^{2}[/itex]<1 is an open subset of (R[itex]^{2}[/itex],d[itex]_{\infty}[/itex]), what does (R[itex]^{2}[/itex],d[itex]_{\infty}[/itex]) mean?
 

Related to Proving Openness in Metric Spaces

1. What is a metric space?

A metric space is a mathematical structure that consists of a set of objects, called points, and a function called a metric that measures the distance between any two points in the set. The metric must satisfy certain properties, such as being non-negative and symmetric.

2. What is an open ball in a metric space?

In a metric space, an open ball is a set of points that are all within a certain distance, called the radius, from a given center point. The open ball is defined as all points that are strictly less than the radius distance from the center point, and it is denoted by B(x, r), where x is the center point and r is the radius.

3. How is an open ball different from a closed ball?

An open ball does not include its boundary points, while a closed ball includes all of its boundary points. In other words, an open ball is a collection of points that are strictly inside the given radius, while a closed ball includes points that are exactly on the boundary.

4. What is the significance of open balls in a metric space?

Open balls are important in metric spaces because they allow us to define neighborhoods around a given point. This concept is useful in many areas of mathematics, including analysis and topology, and it allows us to study the behavior of functions and sets in a local, rather than global, sense.

5. Can an open ball be empty in a metric space?

Yes, an open ball can be empty in a metric space. This occurs when the given center point is not contained in the set, or when the radius is equal to or less than 0. In this case, there are no points within the open ball, and it is considered an empty set.

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