- #1

littleHilbert

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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.

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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