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I suddenly got this doubt, and am a bit confused. Consider the function 1/|x|. As x tends to zero, should i say limit does not exist, or should i say, limit exists but is infinity? Better yet, consider y=x, as x tends to infinity. Rkr1991 ( talk) 07:02, 9 June 2009 (UTC)
Can anyone with adequate knowledge provide some insight about this? — Anonymous Dissident Talk 10:55, 9 June 2009 (UTC)
This is maybe rather a question concerning language than math: When Conway introduced this sequence in his 1986 article, he directly explained the rule and stated: "I note that more usually one is given a sequence such as 55555 ; 55 ; 25 ; 1215 ; 11121115 and asked to guess the generating rule or next term." Can we conclude from that whether he invented the system himself or not? If not - is there a way to ask him (there is no public contact address)? -- KnightMove ( talk) 12:20, 9 June 2009 (UTC)
Hi. I'm studying for a complex analysis qualifying exam, and I'm working on a question about harmonic functions:
Find a harmonic function h on the upper half-plane such that on the positive real axis and on the negative real axis.
So, I figure that, given , the function is harmonic, and it equals 1 on the line Re w = 1, and -1 on the line Re w = -1. All I need is a function that takes the upper half-plane to that strip, sending the appropriate parts of the boundary to the right places. I found a function that pretty much does that. I start with the upper half-plane, apply the principal logarithm to get the strip . Then we just multiply by , and then subtract 1, and finally multiply by -1. This ought to send the upper half-plane to the strip between x=-1 and x=1. The whole function, composed together, comes out to: . Our function h should be the real part of w, and we'll let z = x + iy.
Thus, . This function is harmonic on the upper half-plane (I checked its derivatives), but it's not defined on the real axis. Thus, do I want a function defined piecewise, that is h(x,y) when y>0; -1 when y=0, x<0; and 1 when y=0, x>0? Is that legit? Is it still a harmonic function, thus extended? - GTBacchus( talk) 16:46, 9 June 2009 (UTC)
Hi there,
I've just shown why the set of points in with 1 or 2 rational coordinates is path connected - but I was wondering, what happens if we remove too though? Is the space still connected? Are there even any non-trivial open sets in this space? It seems like any open set would contain a rational point if I'm not being stupid, which would force the indiscrete topology and thus connectedness on this set, right?
Thanks, 131.111.8.97 ( talk) 17:13, 9 June 2009 (UTC)
Consider a simpler situation; for instance, that of the topology of the unit interval. Intuitively, a topology provides a vague notion of "nearness"; that is, using the open sets (members of the topology), one can speak of whether x is close to y but in general, no concrete distance can be obtained. If one considers [0,1], which points are close to 1/2 and which sets are required to convey this? Notice that sets of the form (0.5 - ε, 0.5 + ε) for ε > 0, convey all information about which points are close to 0.5. If a point is close to 0.5, it must be in one of these sets for small enough ε; the smaller ε is, the closer the point is to 0.5. On the other hand, which points are close to 1? Well, since we are studying [0,1], we cannot consider points greater than 1 because they do not belong to [0,1]. Thus, the system of sets conveying "nearness" information is (1 - ε, 1] for ε > 0. One (1) is included in every such set because 1 is the point closest to 1 among all elements of [0,1], trivially. Equivalently, all such sets are the intersection of an open set of the form (1 - ε, 1 + ε) with [0,1]. Generalizing this, we arrive at the definition of the subspace topology. In particular, subsets of R (or R2) are "isolated" from R topologically. In conveying nearness information within these sets, we cannot include points outside these sets.
With respect to your example, we call a set connected if is the union of two disjoint, non-empty open sets. Notice that connectedness should not depend on how it is embedded in Euclidean space; it is a property completely determined by the nearness information within that set. For example, a circle is intuitively connected, and no matter how we twist and turn it, it will still be connected. It does not matter if we are twisting it within the plane or in 3-space because connectedness is a inherent property. In the context of your example (the first one), the space is connected. Why? Intuitively, one has a high degree of freedom since the rationals are countable. Every point in the plane has uncountably many lines going through it (of different slopes). At least one of these lines will not contain any point with rational co-ordinates, if the point through which the line passes has at least one irrational co-ordinate. This is because we would otherwise notice that any such line can be associated to a point in in a one-one manner contradicting the countability of the rationals. Using this line, one can establish that the space is path connected as someone mentioned above.
As another example, consider the plane with an infinite line removed. It is intuitively clear that this set is disconnected. Were we to allow open subsets of the plane to be open in this set, it would be connected since the plane is connected. This is clearly wrong, not only for the obvious reason but also because open subsets of this set have to be subsets of this set and not all open subsets of the plane will have such a property (for instance, any disk intersecting the removed line). In this example, the set of points to "one side of the line" and the set of points to "the other side" provides a separation of the set.
Hope this helps. -- PS T 02:27, 10 June 2009 (UTC)
I had not noticed the above post by pma before I saw this thread! Pma, is it that you are looking for a countable Hausdorff space that is connected (of course, with more than one point)? I have not read all the posts above, and it seems that someone has already proved this but anyway (in these proofs, I am aware of the different definitions mathematicians give to "Hausdorff" etc..., but I always assume that one-point sets are closed):
1. A connected normal space/regular space, X, with more than one point must be uncountable.
Proof
I think that what EmilJ says above accounts for this, but I am including this "different" proof anyway. If x and y are distinct points in X, we apply Urysohn's lemma to deduce that there is a continuous function from X into the unit interval, attaining 0 at x, and 1 at y (in fact, normality is too strong a requirement here). By the intermediate value theorem, the image of X under this continuous function is the entire unit interval. I think it now follows that X is uncountable thus arriving at a contradiction.
Let X be a regular connected space with more than one point. If X were countable, X would be Lindelof and thus normal (every regular Lindelof space is normal; suprisingly, this was unnoticed by Urysohn when he proved Urysohn's metrization theorem!). The theorem now follows from the previous paragraph.
2. There does exist a connected Hausdorff space with more than one point that is countable. However, there cannot exist such a completely Hausdorff space by the previous theorem (notice that normality was too strong).
Proof
Let X = {(x, y) | y ≥ 0 and (x, y) ∈ Q x Q}. Let i be any irrational number. The topology on X is generated by the following basis (Q is the set of all rational numbers):
For every (x, y) in X, let Bε(x, y) = {(x, y)} U Nε(x + y/i) U Nε(x - y/i) where Nε(r) = {q | q ∈ Q and |q - r| < ε.
Then X is a countable connected Hausdorff space under this topology. I think the details behind this are not too hard to see (the irrationality of i is important here). Furthermore, notice that X is not path connected or regular/normal.
Note that, when I was trying to find such a counterexample, I was thinking along the lines of "there has to be points whose every neighbourhood intersects some set to ensure connectivity (in this case, the interesection of the closures of any two non-empty open sets is non-empty); however, these neighbourhoods should "shrink" ensuring the Hausdorff condition." Although this is the case, I do not claim originality of this topology; it appears to be well-known after some research in the literature. This topology is called the " irrational slope topology" because intuitively, neighbourhoods of a point are merely that point, along with intervals about its projection along a line of slope +i and -i into the x-axis. Unfortunately, there is no WP article on the concept. P.S. There may be mistakes in the above argument but I will see if I can do some research and get the exact version of the topology. -- PS T 02:03, 12 June 2009 (UTC)
I came up with a different example meanwhile. Let be an enumeration of all pairs of distinct natural numbers, and let 2 < p0 < p1 < p2 < … be a sequence of primes such that . Put for k ∈ ω and α = 0, 1. Let X be the topological space with underlying set ω and subbasis . Note that X has a basis consisting of sets of the form where k1 < k2 < … < kr, and each such set is nonempty by the Chinese remainder theorem. The closure of is . The space is obviously Hausdorff by the construction. With a little more help of the Chinese remainder theorem, one can show that the intersection of the closures of any two (or finitely many for that matter) nonempty open sets is nonempty, hence the space is connected. — Emil J. 09:59, 12 June 2009 (UTC)
Mathematics desk | ||
---|---|---|
< June 8 | << May | June | Jul >> | June 10 > |
Welcome to the Wikipedia Mathematics Reference Desk Archives |
---|
The page you are currently viewing is an archive page. While you can leave answers for any questions shown below, please ask new questions on one of the current reference desk pages. |
I suddenly got this doubt, and am a bit confused. Consider the function 1/|x|. As x tends to zero, should i say limit does not exist, or should i say, limit exists but is infinity? Better yet, consider y=x, as x tends to infinity. Rkr1991 ( talk) 07:02, 9 June 2009 (UTC)
Can anyone with adequate knowledge provide some insight about this? — Anonymous Dissident Talk 10:55, 9 June 2009 (UTC)
This is maybe rather a question concerning language than math: When Conway introduced this sequence in his 1986 article, he directly explained the rule and stated: "I note that more usually one is given a sequence such as 55555 ; 55 ; 25 ; 1215 ; 11121115 and asked to guess the generating rule or next term." Can we conclude from that whether he invented the system himself or not? If not - is there a way to ask him (there is no public contact address)? -- KnightMove ( talk) 12:20, 9 June 2009 (UTC)
Hi. I'm studying for a complex analysis qualifying exam, and I'm working on a question about harmonic functions:
Find a harmonic function h on the upper half-plane such that on the positive real axis and on the negative real axis.
So, I figure that, given , the function is harmonic, and it equals 1 on the line Re w = 1, and -1 on the line Re w = -1. All I need is a function that takes the upper half-plane to that strip, sending the appropriate parts of the boundary to the right places. I found a function that pretty much does that. I start with the upper half-plane, apply the principal logarithm to get the strip . Then we just multiply by , and then subtract 1, and finally multiply by -1. This ought to send the upper half-plane to the strip between x=-1 and x=1. The whole function, composed together, comes out to: . Our function h should be the real part of w, and we'll let z = x + iy.
Thus, . This function is harmonic on the upper half-plane (I checked its derivatives), but it's not defined on the real axis. Thus, do I want a function defined piecewise, that is h(x,y) when y>0; -1 when y=0, x<0; and 1 when y=0, x>0? Is that legit? Is it still a harmonic function, thus extended? - GTBacchus( talk) 16:46, 9 June 2009 (UTC)
Hi there,
I've just shown why the set of points in with 1 or 2 rational coordinates is path connected - but I was wondering, what happens if we remove too though? Is the space still connected? Are there even any non-trivial open sets in this space? It seems like any open set would contain a rational point if I'm not being stupid, which would force the indiscrete topology and thus connectedness on this set, right?
Thanks, 131.111.8.97 ( talk) 17:13, 9 June 2009 (UTC)
Consider a simpler situation; for instance, that of the topology of the unit interval. Intuitively, a topology provides a vague notion of "nearness"; that is, using the open sets (members of the topology), one can speak of whether x is close to y but in general, no concrete distance can be obtained. If one considers [0,1], which points are close to 1/2 and which sets are required to convey this? Notice that sets of the form (0.5 - ε, 0.5 + ε) for ε > 0, convey all information about which points are close to 0.5. If a point is close to 0.5, it must be in one of these sets for small enough ε; the smaller ε is, the closer the point is to 0.5. On the other hand, which points are close to 1? Well, since we are studying [0,1], we cannot consider points greater than 1 because they do not belong to [0,1]. Thus, the system of sets conveying "nearness" information is (1 - ε, 1] for ε > 0. One (1) is included in every such set because 1 is the point closest to 1 among all elements of [0,1], trivially. Equivalently, all such sets are the intersection of an open set of the form (1 - ε, 1 + ε) with [0,1]. Generalizing this, we arrive at the definition of the subspace topology. In particular, subsets of R (or R2) are "isolated" from R topologically. In conveying nearness information within these sets, we cannot include points outside these sets.
With respect to your example, we call a set connected if is the union of two disjoint, non-empty open sets. Notice that connectedness should not depend on how it is embedded in Euclidean space; it is a property completely determined by the nearness information within that set. For example, a circle is intuitively connected, and no matter how we twist and turn it, it will still be connected. It does not matter if we are twisting it within the plane or in 3-space because connectedness is a inherent property. In the context of your example (the first one), the space is connected. Why? Intuitively, one has a high degree of freedom since the rationals are countable. Every point in the plane has uncountably many lines going through it (of different slopes). At least one of these lines will not contain any point with rational co-ordinates, if the point through which the line passes has at least one irrational co-ordinate. This is because we would otherwise notice that any such line can be associated to a point in in a one-one manner contradicting the countability of the rationals. Using this line, one can establish that the space is path connected as someone mentioned above.
As another example, consider the plane with an infinite line removed. It is intuitively clear that this set is disconnected. Were we to allow open subsets of the plane to be open in this set, it would be connected since the plane is connected. This is clearly wrong, not only for the obvious reason but also because open subsets of this set have to be subsets of this set and not all open subsets of the plane will have such a property (for instance, any disk intersecting the removed line). In this example, the set of points to "one side of the line" and the set of points to "the other side" provides a separation of the set.
Hope this helps. -- PS T 02:27, 10 June 2009 (UTC)
I had not noticed the above post by pma before I saw this thread! Pma, is it that you are looking for a countable Hausdorff space that is connected (of course, with more than one point)? I have not read all the posts above, and it seems that someone has already proved this but anyway (in these proofs, I am aware of the different definitions mathematicians give to "Hausdorff" etc..., but I always assume that one-point sets are closed):
1. A connected normal space/regular space, X, with more than one point must be uncountable.
Proof
I think that what EmilJ says above accounts for this, but I am including this "different" proof anyway. If x and y are distinct points in X, we apply Urysohn's lemma to deduce that there is a continuous function from X into the unit interval, attaining 0 at x, and 1 at y (in fact, normality is too strong a requirement here). By the intermediate value theorem, the image of X under this continuous function is the entire unit interval. I think it now follows that X is uncountable thus arriving at a contradiction.
Let X be a regular connected space with more than one point. If X were countable, X would be Lindelof and thus normal (every regular Lindelof space is normal; suprisingly, this was unnoticed by Urysohn when he proved Urysohn's metrization theorem!). The theorem now follows from the previous paragraph.
2. There does exist a connected Hausdorff space with more than one point that is countable. However, there cannot exist such a completely Hausdorff space by the previous theorem (notice that normality was too strong).
Proof
Let X = {(x, y) | y ≥ 0 and (x, y) ∈ Q x Q}. Let i be any irrational number. The topology on X is generated by the following basis (Q is the set of all rational numbers):
For every (x, y) in X, let Bε(x, y) = {(x, y)} U Nε(x + y/i) U Nε(x - y/i) where Nε(r) = {q | q ∈ Q and |q - r| < ε.
Then X is a countable connected Hausdorff space under this topology. I think the details behind this are not too hard to see (the irrationality of i is important here). Furthermore, notice that X is not path connected or regular/normal.
Note that, when I was trying to find such a counterexample, I was thinking along the lines of "there has to be points whose every neighbourhood intersects some set to ensure connectivity (in this case, the interesection of the closures of any two non-empty open sets is non-empty); however, these neighbourhoods should "shrink" ensuring the Hausdorff condition." Although this is the case, I do not claim originality of this topology; it appears to be well-known after some research in the literature. This topology is called the " irrational slope topology" because intuitively, neighbourhoods of a point are merely that point, along with intervals about its projection along a line of slope +i and -i into the x-axis. Unfortunately, there is no WP article on the concept. P.S. There may be mistakes in the above argument but I will see if I can do some research and get the exact version of the topology. -- PS T 02:03, 12 June 2009 (UTC)
I came up with a different example meanwhile. Let be an enumeration of all pairs of distinct natural numbers, and let 2 < p0 < p1 < p2 < … be a sequence of primes such that . Put for k ∈ ω and α = 0, 1. Let X be the topological space with underlying set ω and subbasis . Note that X has a basis consisting of sets of the form where k1 < k2 < … < kr, and each such set is nonempty by the Chinese remainder theorem. The closure of is . The space is obviously Hausdorff by the construction. With a little more help of the Chinese remainder theorem, one can show that the intersection of the closures of any two (or finitely many for that matter) nonempty open sets is nonempty, hence the space is connected. — Emil J. 09:59, 12 June 2009 (UTC)