Conversely, if I∉U,this implies that the complement of I
the slash is used for the setminus operation. I think using \setminus there (which generates a backslash) would be a more standard notation less likely to be mistaken for quotient structures.
I’m familiar with \setminus being used to denote set complements, so \not\in seemed more appropriate to me (I is not an element of U). I interpret I∖U as “the elements of I not in U,” which is the empty set in this case? (also the elements of U are sets of naturals while the elements of I are naturals, so it’s unclear to me how much this makes sense)
I have heard of filters and ultrafilters, but I have never heard of anyone calling any sort of filter a hyperfilter. Perhaps it is because the ultrafilters are used to make fields of hyperreal numbers, so we can blame this on the terminology. Similarly, the uniform spaces where the hyperspace is complete are called supercomplete instead of hypercomplete.
But the reason why we need to use a filter instead of a collection of sets is that we need to obtain an equivalence relation.
Suppose that I is an index set and Xi is a set with |Xi|>2 for i∈I. Then let M be a collection of subsets of I. Define a relation R on ∏i∈IXi by setting ((xi)i∈I,(yi)i∈I)∈R if and only if {i∈I:xi=yi}∈M. Then in order for R to be an equivalence relation, R must be reflexive, symmetric, and transitive. Observe that R is always symmetric, and R is reflexive precisely when I∈M.
Proposition: The relation R is transitive if and only if M is a filter.
Proof:
← Suppose that M is a filter. Then whenever ((xi)i∈I,(yi)i∈I),((yi)i∈I,(zi)i∈I)∈R, we have
{i∈I:xi=yi},{i∈I:yi=zi}∈M, so since
{i∈I:xi=zi}⊇{i∈I:xi=yi}∩{i∈I:yi=zi}, we conclude that {i∈I:xi=zi} as well. Therefore, ((xi)i∈I,(zi)i∈I)∈R.
→. Suppose now that R,S∈M. Then let let y=0,x=χRc,z=2⋅χSc where χ denotes the characteristic function. Then [x=y]=R,[y=z]=S and [x=z]=R∩S. Therefore,(x,y),(y,z)∈R, so by transitivity, (x,z)∈R as well, hence R∩S=[x=z]∈M.
Suppose now that R⊆S and R∈M. Let x=0,y=χRc and set z=2⋅χSc.
Observe that [x=y]=R and [y=z]=R. Therefore, (x,y),(y,z)∈R. Thus, by transitivity, we know that (x,z)∈R. Therefore, S=[x=z]∈M. We conclude that M is closed under taking supersets. Therefore, M is a filter.
I have heard of filters and ultrafilters, but I have never heard of anyone calling any sort of filter a hyperfilter.
Oops, my bad. I re-read the post as I was typing to make sure I hadn’t missed any explanation. That can sometimes cause me to type what I read instead of what I intended. I probably interverted the prefixes because they feel similar.
Thank you for the math. I am not sure everything is right with your notations in the second half, it seems to me there must be a typo either for the intersection case or the superset one. But the ideas are clear enough to let me complete the proof.
Thank you for this. it looks like a good first contact with hyperreals.
Two nitpicks:
Ω=(1,2,3,ldots). --> I think you forgot a “\” here and it is messing your formatting up.
It is not clear in the post why we use a hyperfilter, rather than just the set of all infinite sets.
Furthermore after
the slash is used for the setminus operation. I think using \setminus there (which generates a backslash) would be a more standard notation less likely to be mistaken for quotient structures.
I’m familiar with \setminus being used to denote set complements, so \not\in seemed more appropriate to me (I is not an element of U). I interpret I∖U as “the elements of I not in U,” which is the empty set in this case? (also the elements of U are sets of naturals while the elements of I are naturals, so it’s unclear to me how much this makes sense)
Sorry, I was quoting the only parts of the sentence.
What I meant was that I would change
to
I have heard of filters and ultrafilters, but I have never heard of anyone calling any sort of filter a hyperfilter. Perhaps it is because the ultrafilters are used to make fields of hyperreal numbers, so we can blame this on the terminology. Similarly, the uniform spaces where the hyperspace is complete are called supercomplete instead of hypercomplete.
But the reason why we need to use a filter instead of a collection of sets is that we need to obtain an equivalence relation.
Suppose that I is an index set and Xi is a set with |Xi|>2 for i∈I. Then let M be a collection of subsets of I. Define a relation R on ∏i∈IXi by setting ((xi)i∈I,(yi)i∈I)∈R if and only if {i∈I:xi=yi}∈M. Then in order for R to be an equivalence relation, R must be reflexive, symmetric, and transitive. Observe that R is always symmetric, and R is reflexive precisely when I∈M.
Proposition: The relation R is transitive if and only if M is a filter.
Proof:
← Suppose that M is a filter. Then whenever ((xi)i∈I,(yi)i∈I),((yi)i∈I,(zi)i∈I)∈R, we have
{i∈I:xi=yi},{i∈I:yi=zi}∈M, so since
{i∈I:xi=zi}⊇{i∈I:xi=yi}∩{i∈I:yi=zi}, we conclude that {i∈I:xi=zi} as well. Therefore, ((xi)i∈I,(zi)i∈I)∈R.
→. Suppose now that R,S∈M. Then let let y=0,x=χRc,z=2⋅χSc where χ denotes the characteristic function. Then [x=y]=R,[y=z]=S and [x=z]=R∩S. Therefore,(x,y),(y,z)∈R, so by transitivity, (x,z)∈R as well, hence R∩S=[x=z]∈M.
Suppose now that R⊆S and R∈M. Let x=0,y=χRc and set z=2⋅χSc.
Observe that [x=y]=R and [y=z]=R. Therefore, (x,y),(y,z)∈R. Thus, by transitivity, we know that (x,z)∈R. Therefore, S=[x=z]∈M. We conclude that M is closed under taking supersets. Therefore, M is a filter.
Q.E.D.
Oops, my bad. I re-read the post as I was typing to make sure I hadn’t missed any explanation. That can sometimes cause me to type what I read instead of what I intended. I probably interverted the prefixes because they feel similar.
Thank you for the math. I am not sure everything is right with your notations in the second half, it seems to me there must be a typo either for the intersection case or the superset one. But the ideas are clear enough to let me complete the proof.