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Bounded sets

  • #1
I am a bit disturbed by the following elementary observation.

Let [tex](X,d)[/tex] be a metric space and [tex]\emptyset\neq A \subseteq X[/tex].
(a) The diameter [tex]\delta (A)[/tex] of [tex]A[/tex] is defined to be [tex]\delta (A):=\sup_{(x,y) \in A^2}d(x,y)[/tex], where [tex]A^2:=A \times A[/tex]
(b) [tex]A[/tex] is called bounded if [tex]\delta (A)<\infty[/tex].

Now let [tex]A,B[/tex] be two nonempty bounded subsets of [tex]X[/tex].
Show that their union [tex]A \cup B[/tex] is bounded.

Let us take the special case [tex]A,B\subset\mathbb{R}[/tex] with the usual distance [tex]d(x,y):=|x-y|[/tex] for [tex]x,y\in\mathbb{R}[/tex]. If we take the ordinary definition of boundedness, then the condition is that [tex]\exists M_1>0\; \forall a\in A : d(a,0)\le M_1[/tex] and [tex]\exists M_2>0\; \forall b\in B : d(b,0)\le M_2[/tex]. Then the usual argument goes by defining the constant [tex]K:=\max\{M_1,M_2\}[/tex], so that in fact [tex]\forall x\in A \cup B:d(x,0)\le K[/tex].

Now if we take the general case and use the above definitions for boundedness via supremum over pairs of points, then it seems that while defining the constant [tex]K:=\max\{\sup_{A^2}d(x,y),\sup_{B^2}d(x,y)\}[/tex] we overlook the case that the sets [tex]A[/tex] and [tex]B[/tex] are disjoint, so that when forming the supremum over the union we have measured the distance between a point [tex]x\in A[/tex] and [tex]y\in B[/tex], and thus brought the distance between the sets [tex]A[/tex] and [tex]B[/tex] into play, which is defined by [tex]D(A,B):=\inf_{(x,y)\in A \times B}d(x,y)[/tex] and should not be neglected.

Wouldn't it then be necessary to take for [tex]K[/tex] the quantity [tex]K:=\sup_{A^2}d(x,y)+\sup_{B^2}d(x,y)+\inf_{A \times B}d(x,y)[/tex], so that [tex]\sup_{A \cup B}d(x,y)\le K[/tex]?

I can't believe that the above definition of boundedness is not equivalent to the ordinary definition as we use it on the real line. Please correct me if there is in fact a misunderstanding on my part.
 
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Answers and Replies

  • #2
Fredrik
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You can prove that the two definitions of "bounded" are equivalent by proving these two statements:

1. Suppose A is a subset of a metric space. Then A has finite diameter if and only if A is a subset of some open ball.

2. Suppose A is a subset of the real numbers. Then A is a subset of some open ball if and only if it's a subset of some open ball around 0.

I'm not sure why you think it matters that the particular "K" you defined isn't always greater than the diameter of [itex]A\cup B[/itex].
 
  • #3
1. Suppose A is a subset of a metric space. Then A has finite diameter if and only if A is a subset of some open ball.
Suppose [tex]\delta (A)<\infty[/tex]. Let [tex]x\in A[/tex] be arbitrary but fixed. Then [tex]d(x,y)\le\sup_{(x,y)\in\{x\}\times A}d(x,y)\le\delta (A)[/tex]. Thus setting [tex]r:=\delta (A)[/tex] we find a (closed) ball [tex]\bar{B}_r(x):=\{y\in X:d(x,y)\le r\}[/tex]. It contains all points of [tex]A[/tex], since for all [tex]x\in A[/tex] and all [tex]y\in\bar{B}_r(x)[/tex] we have by definition [tex]\delta (A)\ge d(x,y)[/tex]

Conversely, suppose there exists a ball of finite radius such that [tex]A\subseteq\bar{B}_r(x)[/tex]. Take any [tex]x\in A[/tex]. For all [tex]y\in\bar{B}_r(x):d(x,y)\le r[/tex]. So [tex]r\ge\sup_{(x,y)\in\{x\}\times\bar{B}_r(x)}d(x,y)\ge\sup_{(x,y)\in\{x\}\times A}d(x,y)[/tex]. Since the left-hand side is independent of the particular choice of [tex]x\in A[/tex], we can take supremum over [tex]A[/tex] for [tex]x[/tex] to obtain

[tex]\sup_{A^2}d(x,y)\le r<\infty[/tex].

I'm not sure why you think it matters that the particular "K" you defined isn't always greater than the diameter of .
Because a finite value of K could be used to bound the diameter from above, and I thought that [tex]
K:=\max\{\sup_{A^2}d(x,y),\sup_{B^2}d(x,y)\}
[/tex] is not large enough to bound the diameter.

Essentially I was thinking about this:

On [tex]\mathbb{R}[/tex] take any numbers [tex]a<b<c<d[/tex] and consider the intervals [tex](a,b)[/tex] and [tex](c,d)[/tex] as well as their union [tex](a,b)\cup (c,d)[/tex]. Then [tex]\sup_{(a,b)^2}d(x,y)=|b-a|,\sup_{(c,d)^2}d(x,y)=|d-c|[/tex] and
[tex]\sup_{[(a,b)\cup (c,d)]^2}d(x,y)=|d-a|\le|d-c|+|c-b|+|b-a|[/tex]. In particular for two points

[tex]\frac{a+b}{2}\in(a,b)[/tex] and [tex]\frac{c+d}{2}\in(c,d)[/tex]

their distance [tex]\frac{c+d}{2}-\frac{a+b}{2}\le|c-b|+\frac{1}{2}(|b-a|+|d-c|)\le|c-b|+\max\{|b-a|,|d-c|\}[/tex]. So unless the distance [tex]|c-b|[/tex] between the intervals is zero, it seems one cannot bound the diameter of the union only by the maximum of the diameters of the constituting intervals. And isn't it what we are required to do if we formulate the boundedness condition in terms of balls?
 
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  • #4
Fredrik
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Suppose e.g. that we're dealing with the real numbers, and that A is an open ball of radius 1 around 0, and B is an open ball of radius 1 around 100. Then your K is going to be =2, and the diameter of [itex]A\cup B[/itex] is going to be 102. You don't have to show that the diameter of [itex]A\cup B[/itex] is ≤2. You just have to show that it's finite.
 
  • #5
You just have to show that it's finite.
Well, I realised this at the very beginning, but it's now that I can write it down:

Let [tex]A[/tex] and [tex]B[/tex] be bounded. W.l.o.g assume that [tex]\delta (A)\le\delta (B)[/tex]. By assumption there are numbers [tex]K_A[/tex] and [tex]K_B[/tex] such that [tex]\delta(A)\le K_A[/tex] and [tex]\delta(B)\le K_B[/tex]. Take [tex]K:=\max\{K_A,K_B\}<\infty[/tex] (i.e. we take a ball that contains both sets A and B). Then for all [tex]x,y\in A \cup B[/tex] we have [tex]d(x,y)\le K[/tex]. Thus [tex]\sup_{x,y\in A \cup B}\le K<\infty[/tex].

Correct?
 
  • #6
Fredrik
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No, that's not correct. Look at the example I mentioned in my previous post. You will have KA=KB=2, and therefore K=2, but the diameter of [itex]A\cup B[/itex] is 102.

I would use the fact that a set is bounded if and only if it's a subset of an open ball. If

[tex]A\subset B_r(p),\ B\subset B_s(q)[/tex]

you can e.g. define t=s+d(p,q) and prove that

[tex]A\cup B\subset B_t(p)[/tex]

If you're going to use the equivalent definition of "bounded", the one involving a supremum, you need to say something about the distance between the two sets. (You were right about that part in your first post).
 
  • #7
I should have made it more explicit. Here is the complete version:

Let [tex]A[/tex] and [tex]B[/tex] be bounded. Take any point [tex]z_0\in X[/tex], such that for all [tex]x\in A[/tex] we have [tex]d(x,z_0)\ge\sup_{u\in A}d(x,u)[/tex] and for all [tex]y\in B[/tex] respectively [tex]d(y,z_0)\ge\sup_{v\in B}d(y,v)[/tex]. Then with [tex]K_A:=\sup_{x\in A}d(x,z_0)[/tex] and [tex]K_B:=\sup_{y\in B}d(y,z_0)[/tex] we have that [tex]\delta(A)\le K_A[/tex] and [tex]\delta(B)\le K_B[/tex]. Take [tex]K:=2\max\{K_A,K_B\}<\infty[/tex] (i.e. we take a ball that contains both sets A and B). Then for all [tex]x,y\in A \cup B[/tex] we have [tex]d(x,y)\le d(x,z_0)+d(z_0,y)\le K_A+K_B\le K[/tex]. Thus [tex]\sup_{x,y\in A \cup B}d(x,y)\le K<\infty[/tex].

Again that's all about balls around the same "reference point" like e.g 0 on the real line.
 
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  • #8
Fredrik
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Still wrong. (And also weird and awkward). Your [itex]z_0[/itex] is any point that's not in [itex]A\cup B[/itex]. Look at my example again, and choose [itex]z_0=50[/itex]. Now [itex]K_A=K_B=51[/itex], so K=51 and d(0,100)=100>51, even though 0 is in A and 100 is in B.

I don't know what "w.l.o.g" means, and I don't understand your last comment, since you didn't actually use any balls in your proof. You just declared your intention to use one, and then didn't.
 
  • #9
Oh my it's getting really funny! Yeah, but the fact is that both sets are in the ball with radius 51, and so their union is bounded. I overlooked the value of the constant K, which should of course be the diameter of the (larger) ball, i.e. 102.

I "repaired" the proof. It should be OK now, hopefully! :))

W.l.o.g. means "without loss of generality". But I see that the way the diameters of the two sets relate to each other does not really matter here, so I deleted the corresponding sentence.
 
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