Uncountably many vertical asymptotes?
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Uncountably many vertical asymptotes?
My friend and I were on the porch having a smoke and discussing mathematical oddities when he brought up vertical asymptotes, leading later to questions which I pose as follows:
Consider a function [imath]f : E \to \mathbb{R}[/imath], where [imath]E \subset \mathbb{R}[/imath]. Let A be the set of points at which the graph of f has a vertical asymptote.
1) Does there exist such an f such that A is uncountable?
2) Does there exist such an f such that A is dense in E?
As cool a pathology as something fulfilling 2 would be, I believe I have proven no such f exists, but I would like to know if my proof has an error: Suppose such an f exists. Let [imath]M > 0[/imath] be given. For each [imath]a \in A[/imath], choose a [imath]\delta (a) > 0[/imath] such that whenever [imath]x \in E \cap \lbrack a  \delta (a),\ a + \delta (a) \rbrack[/imath], we have [imath]f(x) > M[/imath]. Since A is dense in E, the union of all [imath]\lbrack a  \delta (a),\ a + \delta (a) \rbrack[/imath] contains all of E. Thus [imath]f(x) > M[/imath] for any M on its entire domain, so f is defined nowhere.
I feel like I may be unjustified in claiming that a collection of closed intervals centered at points dense in a set of reals will cover the set, and also that there should be an easier way to prove this. Regarding question 1, though, I don't know where I would go about starting a proof; for that matter, I imagine it's not at all unlikely that such functions exist. In that case A would have to have a limit point, but functions like tan(1/x) demonstrate that that's no big deal. Any thoughts?
Consider a function [imath]f : E \to \mathbb{R}[/imath], where [imath]E \subset \mathbb{R}[/imath]. Let A be the set of points at which the graph of f has a vertical asymptote.
1) Does there exist such an f such that A is uncountable?
2) Does there exist such an f such that A is dense in E?
As cool a pathology as something fulfilling 2 would be, I believe I have proven no such f exists, but I would like to know if my proof has an error: Suppose such an f exists. Let [imath]M > 0[/imath] be given. For each [imath]a \in A[/imath], choose a [imath]\delta (a) > 0[/imath] such that whenever [imath]x \in E \cap \lbrack a  \delta (a),\ a + \delta (a) \rbrack[/imath], we have [imath]f(x) > M[/imath]. Since A is dense in E, the union of all [imath]\lbrack a  \delta (a),\ a + \delta (a) \rbrack[/imath] contains all of E. Thus [imath]f(x) > M[/imath] for any M on its entire domain, so f is defined nowhere.
I feel like I may be unjustified in claiming that a collection of closed intervals centered at points dense in a set of reals will cover the set, and also that there should be an easier way to prove this. Regarding question 1, though, I don't know where I would go about starting a proof; for that matter, I imagine it's not at all unlikely that such functions exist. In that case A would have to have a limit point, but functions like tan(1/x) demonstrate that that's no big deal. Any thoughts?
Re: Uncountably many vertical asymptotes?
Let {r_n} be an enumeration of Q which is dense in R. Suppose it works out that delta(r_n) = 1/2^n; then the measure of the union of all the [r_n  delta(r_n), r_n + delta(r_n)] is at most 2, so the union of the intervals can't be all of R.
Re: Uncountably many vertical asymptotes?
I'm sure you can "drape" some sort of function over the cantor set, which is uncountable. That is, every time you remove a third, put a suitable "Ushaped" function over the portion you removed, so that the endpoints of the remaining set are the asymptotes.
As far as it being dense... I'd imagine that if you define "vertical asymptote" a bit more rigorously, you'll have to at least have some sort of open set on one side of the asymptote on which the function is actually defined, and then to be dense, you'd have to have an asymptote inside that open set, and so f isn't defined on the set after all.
EDIT: There is a flaw in your proof... and it's one of those fun things in math that completely destroys any intuition you thought you had about anything. I can place closed, non degenerate intervals over a dense subset of some set E and still not cover E. Take the rationals in [0,1] (call it [imath]Q[/imath]). Now, those rationals are countable, so I have bijective [math]f:Q \rightarrow \mathbb{N}[/math] Now then, for each [imath]q \in Q[/imath], I place over it the interval [math]\left[q  \frac{1}{2^{f(q)+2}}, q + \frac{1}{2^{f(q) + 2}}\right][/math] The length of each interval is thus of length [math]\frac{1}{2^{f(q) + 1} }[/math]. Now, add up all the lengths! That is, find:
[math]\sum_{i=1}^{\infty} \frac{1}{2^{i+1}} = \frac{1}{2}\sum_{i=1}^{\infty} \frac{1}{2^{i}} = \frac{1}{2}[/math]
But if these intervals are supposed to cover [0,1], shouldn't they add up to have length 1?
As far as it being dense... I'd imagine that if you define "vertical asymptote" a bit more rigorously, you'll have to at least have some sort of open set on one side of the asymptote on which the function is actually defined, and then to be dense, you'd have to have an asymptote inside that open set, and so f isn't defined on the set after all.
EDIT: There is a flaw in your proof... and it's one of those fun things in math that completely destroys any intuition you thought you had about anything. I can place closed, non degenerate intervals over a dense subset of some set E and still not cover E. Take the rationals in [0,1] (call it [imath]Q[/imath]). Now, those rationals are countable, so I have bijective [math]f:Q \rightarrow \mathbb{N}[/math] Now then, for each [imath]q \in Q[/imath], I place over it the interval [math]\left[q  \frac{1}{2^{f(q)+2}}, q + \frac{1}{2^{f(q) + 2}}\right][/math] The length of each interval is thus of length [math]\frac{1}{2^{f(q) + 1} }[/math]. Now, add up all the lengths! That is, find:
[math]\sum_{i=1}^{\infty} \frac{1}{2^{i+1}} = \frac{1}{2}\sum_{i=1}^{\infty} \frac{1}{2^{i}} = \frac{1}{2}[/math]
But if these intervals are supposed to cover [0,1], shouldn't they add up to have length 1?
Last edited by z4lis on Sat Mar 21, 2009 4:41 am UTC, edited 1 time in total.
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Re: Uncountably many vertical asymptotes?
Here's the problem: strictly speaking, a function on [imath]E[/imath] is defined everywhere on [imath]E[/imath], so it can't have any "vertical asymptotes." And I can't think of a good way to formalize a generic definition of "vertical asymptote" that isn't trivial. Meromorphic functions are defined to have isolated poles, which are countable and not dense by construction (because meromorphic functions turn out to be nice to study if we make this restriction). Elements of [imath]L^p[/imath] spaces still differ from a function defined everywhere by a set of measure zero. And I'm reasonably certain this question is trivial for distributions if stated appropriately.

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Re: Uncountably many vertical asymptotes?
Ah, I figured I was making too bold a claim there.
Mm, the Cantor set "draping" sounds like a good approach.
As far as rigorously defining "vertical asymptote" on the real line, I was just working with t being a vertical asymptote (from the left) if [imath]\lim_{x \to t^}f(x) = \pm \infty[/imath], and likewise from the right. With this definition, however, f could be defined at an asymptote if it wanted.
Mm, the Cantor set "draping" sounds like a good approach.
As far as rigorously defining "vertical asymptote" on the real line, I was just working with t being a vertical asymptote (from the left) if [imath]\lim_{x \to t^}f(x) = \pm \infty[/imath], and likewise from the right. With this definition, however, f could be defined at an asymptote if it wanted.
Re: Uncountably many vertical asymptotes?
One common counterexample function (integrable but unbounded on every interval) is [imath]2^{n}f(xr_n)[/imath], where [imath]r_n[/imath] is an enumeration of the rationals, and, say, [imath]f(x)=\chi_{[1,1]}(x)x^{1/2}[/imath]. Depending on how you define "vertical asymptote", it'll satisfy your condition 2.
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Re: Uncountably many vertical asymptotes?
Okay, suppose you have uncountably many leftvertical asymptotes, i.e. a set of numbers x_{i} for i \in I (with I uncountable) so that the limit [imath]\lim_{x \rightarrow x_i^} f(x) = \pm \infty[/imath]. For each asymptote at x_{i}, take an open interval (a_i,x_i) that contains no other asymptotes. (There must be such an interval, or else the limit wouldn't exist.) Then you've just found uncountably many disjoint open intervals in R, which is impossible.
Re: Uncountably many vertical asymptotes?
Buttons wrote:Okay, suppose you have uncountably many leftvertical asymptotes, i.e. a set of numbers x_{i} for i \in I (with I uncountable) so that the limit [imath]\lim_{x \rightarrow x_i^} f(x) = \pm \infty[/imath]. For each asymptote at x_{i}, take an open interval (a_i,x_i) that contains no other asymptotes. (There must be such an interval, or else the limit wouldn't exist.) Then you've just found uncountably many disjoint open intervals in R, which is impossible.
covered is a bit strong. You've missed all the [imath]x_i[/imath]'s for one...err well more than one
edit: Your stealth edit skills are strong.
Last edited by Yesila on Sun Mar 22, 2009 3:17 am UTC, edited 1 time in total.
Re: Uncountably many vertical asymptotes?
Your counterexample is no match for my stealth edits.
Re: Uncountably many vertical asymptotes?
Buttons wrote:Then you've just found uncountably many disjoint open intervals in R, which is impossible.
Is it not possible for a "uncountable" series with positive terms to converge?
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Re: Uncountably many vertical asymptotes?
I have no idea what you mean by an uncountable series. But you cannot find an uncountable family of nonempty disjoint open sets in R: each one contains a rational number.
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Re: Uncountably many vertical asymptotes?
z4lis wrote:Buttons wrote:Then you've just found uncountably many disjoint open intervals in R, which is impossible.
Is it not possible for a "uncountable" series with positive terms to converge?
No. If you have uncountably many terms, infinitely many of them must be larger than 1/n for some n (or else you would only have countably many), and then your sum would be at least as large as infinity times 1/n.

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Re: Uncountably many vertical asymptotes?
Something that sort of fits what you're looking for:
f(x) = 0 if x is irrational and if x is rational, say x = m/n, then f(x) = n
f is locally unbounded everywhere. You don't get the limit of f approaching infinity as x approaches a point (since it doesn't exist) but the limsup does for every point. If you just define f over the rationals then you get the limit behavior that you want
f(x) = 0 if x is irrational and if x is rational, say x = m/n, then f(x) = n
f is locally unbounded everywhere. You don't get the limit of f approaching infinity as x approaches a point (since it doesn't exist) but the limsup does for every point. If you just define f over the rationals then you get the limit behavior that you want
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Re: Uncountably many vertical asymptotes?
Buttons wrote:Okay, suppose you have uncountably many leftvertical asymptotes, i.e. a set of numbers x_{i} for i \in I (with I uncountable) so that the limit [imath]\lim_{x \rightarrow x_i^} f(x) = \pm \infty[/imath]. For each asymptote at x_{i}, take an open interval (a_i,x_i) that contains no other asymptotes. (There must be such an interval, or else the limit wouldn't exist.) Then you've just found uncountably many disjoint open intervals in R, which is impossible.
This doesn't work. Let f be defined on [imath][\frac{1}{2^n},\frac{1}{2^{n+1}})[/imath] by the rule [math]f(x)=\frac{1}{x}+\frac{1}{2^{n+1}x+1}[/math] for n=0, 1, 2, 3, ... Then f is defined on [1,0), has vertical asymptotes at 1/2, 1/4, 1/8, ..., 0 (according to the definition you gave), but there is no interval (x,0) that contains no asymptotes.
I suspect your conclusion is correct however.
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Re: Uncountably many vertical asymptotes?
I still hold that if you define [imath]g:[0,1] \rightarrow R[/imath] with:
[math]g(x) = \begin{cases} 1/x \mbox{ for } x \in [0,\frac{1}{2}] \\ 1  x \mbox{ for } x \in (\frac{1}{2}, 1] \end{cases}[/math]
that has asymptotes at 0 and 1 and for any interval [a,b] define:
[math]g_{[a,b]}(x) = g(\frac{xa}{ba})[/math]
that has asymptotes at a and b, you can then take a f(x) on [0,1] minus the Cantor set whose values are determined by those [imath]g_{[a,b]}[/imath], replacing [a,b] with whatever interval was removed along with x when recursively forming the Cantor set. Seems that the resulting function should have an uncountable number of asymptotes, one for each point in the Cantor set.
[math]g(x) = \begin{cases} 1/x \mbox{ for } x \in [0,\frac{1}{2}] \\ 1  x \mbox{ for } x \in (\frac{1}{2}, 1] \end{cases}[/math]
that has asymptotes at 0 and 1 and for any interval [a,b] define:
[math]g_{[a,b]}(x) = g(\frac{xa}{ba})[/math]
that has asymptotes at a and b, you can then take a f(x) on [0,1] minus the Cantor set whose values are determined by those [imath]g_{[a,b]}[/imath], replacing [a,b] with whatever interval was removed along with x when recursively forming the Cantor set. Seems that the resulting function should have an uncountable number of asymptotes, one for each point in the Cantor set.
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Re: Uncountably many vertical asymptotes?
But, with that function, can you say that there is a vertical asymptote at, say, 1/4? It's never at the border of a removed interval... which is where your vertical asymptotes are. But it is in the Cantor set.
And depending on how you define a vertical asymptote, you probably need to have the function to be defined on a (possibly onesided) neighbourhood of 1/4, and no (even onesided) neighbourhood of 1/4 is disjoint from the Cantor set.
And depending on how you define a vertical asymptote, you probably need to have the function to be defined on a (possibly onesided) neighbourhood of 1/4, and no (even onesided) neighbourhood of 1/4 is disjoint from the Cantor set.
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Re: Uncountably many vertical asymptotes?
Okay, I'm thinking there's two reasonable definitions of "vertical asymptote" floating around here:
The first is "f has a vertical asymptote at a iff, as x approaches a, f(x) is unbounded". Then you can make uncountably many vertical asymptotes (via that Cantor set construction, for example), and you can make them dense (antonfire gave an example of this).
The other is "f has a vertical asymptote at a iff, as x approaches a, f(x) approaches infinity". You can still make uncountably many asymptotes, and make them all dense: Let E be the rational numbers. For each rational q, let f(q) = b, where q = a/b expressed in lowest terms. Then you've got vertical asymptotes at every irrational number.
Or did you want your functions to be continuous on their domains? Then you can't make a dense set of asymptotes: Choose some x where f is defined, then f is bounded on a deltaneighborhood of x. That Cantor set construction might still work, though, as long as you make the "minimum value" on each removed subinterval larger and larger. Then as you get close to any point in the Cantor set, even nonendpoints like 1/4, it seems like the values of f would have to approach infinity (since any sequence approaching 1/4 in the complement of the Cantor set would need to enter smaller and smaller removed intervals).
If your definition of asymptote at a requires f to be defined on a neighborhood of a, even a onesided neighborhood, you'll never get the asymptotes to be uncountable or dense ("defined on a neighborhood" kills denseness, and Buttons's proof for uncountability works in this case).
The first is "f has a vertical asymptote at a iff, as x approaches a, f(x) is unbounded". Then you can make uncountably many vertical asymptotes (via that Cantor set construction, for example), and you can make them dense (antonfire gave an example of this).
The other is "f has a vertical asymptote at a iff, as x approaches a, f(x) approaches infinity". You can still make uncountably many asymptotes, and make them all dense: Let E be the rational numbers. For each rational q, let f(q) = b, where q = a/b expressed in lowest terms. Then you've got vertical asymptotes at every irrational number.
Or did you want your functions to be continuous on their domains? Then you can't make a dense set of asymptotes: Choose some x where f is defined, then f is bounded on a deltaneighborhood of x. That Cantor set construction might still work, though, as long as you make the "minimum value" on each removed subinterval larger and larger. Then as you get close to any point in the Cantor set, even nonendpoints like 1/4, it seems like the values of f would have to approach infinity (since any sequence approaching 1/4 in the complement of the Cantor set would need to enter smaller and smaller removed intervals).
If your definition of asymptote at a requires f to be defined on a neighborhood of a, even a onesided neighborhood, you'll never get the asymptotes to be uncountable or dense ("defined on a neighborhood" kills denseness, and Buttons's proof for uncountability works in this case).
Let's have a fervent argument, mostly over semantics, where we all claim the burden of proof is on the other side!
Re: Uncountably many vertical asymptotes?
MartianInvader wrote:If your definition of asymptote at a requires f to be defined on a neighborhood of a, even a onesided neighborhood, you'll never get the asymptotes to be uncountable or dense ("defined on a neighborhood" kills denseness, and Buttons's proof for uncountability works in this case).
I do think we should requires f to be defined on a punctured neighborhood (or a onesided neighborhood) at each asymptote, but skeptical scientist's counterexample to my proof satisfies this property. I agree with him that the result is still probably true, but we haven't proven it yet.
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Re: Uncountably many vertical asymptotes?
Sorry, I was still referring to the "continuous on their domains" case when I said the proof works. If I'm reading skeptical's example right, it's not continuous at 1/2^n.
If you don't require continuity, then you can probably take my example defined on the rationals and assign values to the irrationals in such a way that f is defined everywhere and has vertical asymptotes at every irrational. This might require the axiom of choice or something, though.
Edit: Actually, I guess you couldn't make ALL the irrationals vertical asymptotes. Of the irrationals in [0,1], for example, you'd have to assign uncountably many of them values that stay inside some yinterval, so you could take a sequence whose yvalues converge and then take a subsequence whose xvalues converge, and thus stay bounded. You could say the same about any xinterval, so at the very least you'd be stuck with a dense set of nonasymptote irrationals.
If we're not requiring conitnuity, I suspect we could still make an uncountable number of asymptotes, and probably make them dense.
Second edit: Okay, I'm wrong, and here's why: I'll prove that you cannot have a set of vertical asymptotes which is dense (if f is defined everywhere on an interval).
Suppose f is defined everywhere in some interval I, and has a dense set of vertical asymptotes. Let S_n be the set of points x \in I such that f(x) < n. Since f(x) is defined on all of I, we know I is the union of the S_n. However, there cannot be a sequence in S_n whose limit is a vertical asymptote since such a sequence cannot approach infinity. Since the vertical asymptotes are dense, this means each S_n is nowhere dense. The Baire Category Theorem says that I cannot be a countable union of nowhere dense sets, giving a contradiction.
This still leaves the uncountable case unanswered. I'm not convinced it's impossible (it still seems you could do something clever with the Cantor set).
If you don't require continuity, then you can probably take my example defined on the rationals and assign values to the irrationals in such a way that f is defined everywhere and has vertical asymptotes at every irrational. This might require the axiom of choice or something, though.
Edit: Actually, I guess you couldn't make ALL the irrationals vertical asymptotes. Of the irrationals in [0,1], for example, you'd have to assign uncountably many of them values that stay inside some yinterval, so you could take a sequence whose yvalues converge and then take a subsequence whose xvalues converge, and thus stay bounded. You could say the same about any xinterval, so at the very least you'd be stuck with a dense set of nonasymptote irrationals.
If we're not requiring conitnuity, I suspect we could still make an uncountable number of asymptotes, and probably make them dense.
Second edit: Okay, I'm wrong, and here's why: I'll prove that you cannot have a set of vertical asymptotes which is dense (if f is defined everywhere on an interval).
Suppose f is defined everywhere in some interval I, and has a dense set of vertical asymptotes. Let S_n be the set of points x \in I such that f(x) < n. Since f(x) is defined on all of I, we know I is the union of the S_n. However, there cannot be a sequence in S_n whose limit is a vertical asymptote since such a sequence cannot approach infinity. Since the vertical asymptotes are dense, this means each S_n is nowhere dense. The Baire Category Theorem says that I cannot be a countable union of nowhere dense sets, giving a contradiction.
This still leaves the uncountable case unanswered. I'm not convinced it's impossible (it still seems you could do something clever with the Cantor set).
Let's have a fervent argument, mostly over semantics, where we all claim the burden of proof is on the other side!
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