Wikipedia:Reference desk/Archives/Mathematics/2009 June 9

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June 9

Can a limit be infinite?

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)

Yes, you'd usually consider those limits to be infinity. Evaluating a limit or some other expression to infinity is usually just fine. It's when you start treating infinity as a number that you get into trouble. --Pykk (talk) 08:19, 9 June 2009 (UTC)

In a casual context, saying the limit "goes to infinity" or "is infinity" will be understood by other people correctly. Saying the limit "exists" is sloppy at best but is probably ok.

Formally, you need to be clear about what set (more precisely, metric space or topological space) one is taking the limit in. If you are working in the real numbers, then the function 1/|x| has no limit as x goes to zero (i.e., the limit does not exist). If you are working in the extended real numbers, which are like the real numbers together with plus and minus "infinity", then the function 1/|x| has a limit of positive infinity as x goes to zero (i.e., the limit exists and is positive infinity). Eric. 131.215.159.106 (talk) 08:47, 9 June 2009 (UTC)

the function 1/|x| has a limit of positive infinity as x goes to zero — not quite: there are two limits, one approaching 0 from above which is positive infinity, and the other which approaches from below, which is negative infinity. — Charles Stewart (talk) 10:15, 9 June 2009 (UTC)

Absolute value? -- Meni Rosenfeld (talk) 11:11, 9 June 2009 (UTC)

Ah, yes. I managed to ignore those vertical lines. Quibble withdrawn. — Charles Stewart (talk) 11:37, 9 June 2009 (UTC)

I think it's common to treat limits involving infinity as complete pieces of notation. That is, "$\lim _{x\to a}f(x)=+\infty$" is defined to mean "for every $M\in \mathbb {R}$ there is a $\delta >0$ such that for every $0<|x-a|<\delta$ we have $f(x)>M$", "$\lim _{x\to +\infty }f(x)=+\infty$" means "for every $M\in \mathbb {R}$ there is an $N\in \mathbb {R}$ such that for every $x>N$ we have $f(x)>M$", and so on. You don't have to explicitly consider a topological space in which there is an element "$+\infty$". -- Meni Rosenfeld (talk) 11:11, 9 June 2009 (UTC)

While you don't need to be explicit about what you are doing, you shouldn't really use that kind of notation without knowing how to make it explicit, otherwise you risk making assumptions about how those limits behave that aren't true. For example, you might a limit that you've shown exists (with this broad definition of existence), so you call it L. You then prove that L+a=L+b for some real variables a and b, so you then deduce that a=b. This would be fallacious since, if L is infinite, you cannot subtract it from both sides. --Tango (talk) 16:36, 9 June 2009 (UTC)

I generally understand those written infinite limits as being little more than an abuse of notation. Under the standard epsilon-delta definition of a limit, those limits do not exist. However, a limit can fail to exist without the limit "being infinity". So saying that the limit is infinity can be thought of as being more specific than saying the limit doesn't exist. --COVIZAPIBETEFOKY (talk) 12:24, 9 June 2009 (UTC)

I would add: ...does not exist in R (whereas it exists as limit in the extended real line). In general, the epsilon-delta definition applies to metric spaces and needs a distance. The compactification of R is metrizable and the epsilon-delta definition applies perfectly. Notice that two elementary compactifications of R are commonly used: the one point (or Alexandrov) compactification , which is homeomorphic to S¹, and where the added point is usually denoted $\infty$ (without sign); and the order compactification, which is homeomorphic to the interval [-1,+1], that has two added points (positive and negative infinity) denoted $+\infty$ and $-\infty$. --pma (talk) 13:16, 9 June 2009 (UTC)

As pma notes above, one needs distance to define the epsilon-delta definition. For the interest of the OP, one can define limits in an arbitrary topological space; this is not the same as the epsilon-delta definition but is still appropriate to some extent.

People have already mentioned this above, but just to emphasize, in the real numbers, "a limit is positive infinity" conveys the information that it is unbounded "to the positive side." In basic calculus, one usually just defines the limit to be infinity without going into technical details but this is "imaginary." Basically, the function is just unbounded. --PST 02:37, 10 June 2009 (UTC)

Just to clarify - the pure topological definition of limits is equivalent to the epsilon-delta one if the topological space in question is a metric space. There is more to having an infinite limit than just being unbounded - $f(x)=x\sin(x)$ is unbounded as x tends to infinity, but one wouldn't say it tends to infinity. In order for it to be said to tend to positive infinity it is necessary that, for any $u\in \mathbb {R}$, there exists an $x_{0}\in \mathbb {R}$ such that $f(x)>u,\,\,\forall x>x_{0}$. --Tango (talk) 03:39, 10 June 2009 (UTC)

I did intend the same meaning as that conveyed above but presented the meaning in an informal manner, the correctness of which might have been ambiguous. Nevertheless, thankyou for clarifying my post. --PST 09:20, 10 June 2009 (UTC)

Wedge again

Can anyone with adequate knowledge provide some insight about this? —Anonymous Dissident^Talk 10:55, 9 June 2009 (UTC)

Look-and-say sequence

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)

Conway is current associated with Princeton University, so can almost certainly be contacted there. Our article says (in the Origins section): "It was introduced and analyzed by John Conway in his paper "The Weird and Wonderful Chemistry of Audioactive Decay" published in Eureka 46, 5-18 in 1986." So that would imply he invented it or, at least, was the first to publish something about it. I haven't read the paper, does it not make it clear where the sequence came from? If it doesn't say he got it from someone it probably means it was his own, since academics are generally very careful to correctly attribute things. --Tango (talk) 16:44, 9 June 2009 (UTC)

The next term is 31123115, and then 132112132115. Fun. I think it's an old game, that predates Conway, but I'm not sure. -GTBacchus^(talk) 16:49, 9 June 2009 (UTC)

Its use in psychology and testing may have been novel, but this coding scheme is very old. Huffman coding was published in 1952, which is a significantly more complex code than run-length encoding. Unfortunately, run-length encoding doesn't have a history section... but it predates huffman codes for data compression. Here is an IEEE "optimized implementation" of this coding scheme from 1959... there seem to be plenty more from the early 1940s and 1950s Nimur (talk) 17:58, 9 June 2009 (UTC)

A more detailed explanation: Conway is definitely the first person to have published something of mathematical interest about this sequence. But: Minor intellectual creations like jokes and riddles are often spread worldwide, the creator remaining unknown (e. g. the Zebra Puzzle).

I'm working on a term paper about this topic, which is described as "Conway turned a fun puzzle into a cosmological theory". But did he so, or is the puzzle a daughter product of his work?

The only hint his paper gives, as stated above: "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." Now does this answer the question? Is the quote to be interpreted that Conway knew the puzzle from somewhere else? --KnightMove (talk) 11:15, 10 June 2009 (UTC)

Harmonic function

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 $h=1$ on the positive real axis and $h=-1$ on the negative real axis.

So, I figure that, given $w=u+iv$, the function $u(w)=Rew$ 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 $\{x+iy|0<y<\pi \}$. Then we just multiply by ${\frac {2}{i\pi }}$, 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: $w=1-{\frac {2}{i\pi }}Log(z)$. Our function h should be the real part of w, and we'll let z = x + iy.

Thus, $h(x,y)={\frac {-2}{\pi }}tan^{-1}({\frac {x}{y}})$. 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)

I think everything is fine, but perhaps you have a sign error? You just want the piecewise function to be continuous at the boundary (real axis other than the origin). Any sequence approaching (x,0) for x>0 must have all but finitely many x_n > x/2 > 0, so you get h(x,0)=-2/pi*arctan(+infinity) = -1, and for x < 0, you have x_n < x/2 < 0, so you get h(x,0)=-2/pi*arctan(-infinity) = 1. For harmonic functions, sometimes "boundary values" can be assigned even when the extension is not continuous, but you function is nice enough not to have to fool with this. JackSchmidt (talk) 17:40, 9 June 2009 (UTC)

Thank you. Now I'll be doing great as soon as I can figure out where I made that sign error... :-) -GTBacchus^(talk) 18:16, 9 June 2009 (UTC)

Set of coordinates in R^2 with exactly 1 rational point - connected?

Hi there,

I've just shown why the set of points in $\mathbb {R} ^{2}$ with 1 or 2 rational coordinates is path connected - but I was wondering, what happens if we remove $\mathbb {Q} \times \mathbb {Q}$ 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)

Take any point $\langle a,b\rangle \notin \mathbb {Q} \times \mathbb {Q}$. Assume for instance $a\notin \mathbb {Q}$ (the other case is symmetric). Using two line segments going through $\langle a,{\sqrt {2}}\rangle$, the point is connected by a path to the point $\langle {\sqrt {2}},{\sqrt {2}}\rangle$. We can do this for any point, hence $\mathbb {R} ^{2}-\mathbb {Q} ^{2}$ is path-connected. You seem to be confused about the definition of the subspace topology: if U is any open subset of $\mathbb {R} ^{2}$, then $U-\mathbb {Q} ^{2}$ is an open set in $\mathbb {R} ^{2}-\mathbb {Q} ^{2}$. — Emil J. 17:33, 9 June 2009 (UTC)

Sorry - to clarify, I meant we start with points without 2 irrational coordinates, so at least 1 coordinate is rational, and then remove also points with 2 rational coordinates, so only points with exactly 1 rational coordinate are allowed (staying in $\mathbb {R} ^{2}$ here) - the open sets in $\mathbb {R} ^{2}$ are just discs with no points 'on the edge', aren't they? 131.111.8.97 (talk) 17:42, 9 June 2009 (UTC)

It's totally disconnected (consider lines of nonzero rational slope and rational y-intercept). Algebraist 17:48, 9 June 2009 (UTC)

Ah, that makes sense, but where am I going wrong then in thinking the fact it isn't path connected means it's the union of open sets which must be balls in R^2, but any of these balls must contain a point with 2 rational coordinates? 131.111.8.97 (talk) 17:58, 9 June 2009 (UTC)

You're still confused about the definition of the subspace topology. The open sets of your funny subspace are restrictions to your funny subspace of the open sets of R². For example, the set of points with exactly one co-ordinate rational and both co-ordinates strictly positive is an open set in your space. Algebraist 18:39, 9 June 2009 (UTC)

So it's $\scriptstyle (\mathbb {R} \setminus \mathbb {Q} )\times \mathbb {Q} \,\cup \,\mathbb {Q} \times (\mathbb {R} \setminus \mathbb {Q} )$: the dark squares of a funny chessboard indeed. --pma (talk) 20:07, 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 $\mathbb {Q} \times \mathbb {Q}$ 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. --PST 02:27, 10 June 2009 (UTC)

That makes much more sense, thankyou everyone for the helpful explanations! The time you spent explaining everything was earnestly appreciated. I've been churning my way through a good few more questions and the only one which has me baffled still is whether the rationals $\mathbb {Q}$ are connected under the 2-adic topology. I understand all the definitions comfortably now I think, but I'm having trouble even picturing the open sets let alone contemplating a way to disconnect the space (if indeed there is one)! Can anyone give me a hint or two to get me started? Thanks again! 131.111.8.104 (talk) 05:25, 10 June 2009 (UTC)

Since you are interested in connectedness, I thought that I might give you a few problems on it. These problems may serve to clarify intuition and may be useful should you wish to do more exercises. Since I am not familiar with your exact amount of knowledge, I will attempt to give problems which require only basic definitions. If you have further interest, I recommend the book "Counterexamples in topology"; a fine book with many problems addressing several notions in point-set topology.

1. A function defined on a topological space X is said to be "locally constant" if each point in X has a open neighbourhood on which the function is constant. If X is connected, and f is locally constant on X, must f be constant on X?

2. Is every open subset of the real line a disjoint union of countably many open intervals? Prove or disprove.

3. If on the set of real numbers, one defines U to be open if and only if the complement of U is finite or all of R, is R connected under this topology?

4. A topological space is called "hyperconnected", if there does not exist two disjoint non-empty open sets. Observe that every hyperconnected space is connected. Is the topology in question 3 hyperconnected? Further, prove that a continuous function from a hyperconnected space to a Hausdorff space must be constant.

5. Let f and g be two continuous functions defined on the real line, with range a Hausdorff space. Suppose further that f(x) = g(x) for every rational number x. Prove that f(x) = g(x) for any real number x.

6. Can a countable set be connected under some topology? If so, does there exist an infinite topology on a countable set that is connected?

7. Does there exist a connected set with an infinite topology, every subspace of which is connected?

If you have any questions on the above problems feel free to ask them here. If you find them easy (which hopefully you should after some experience with connectedness!), don't worry too much about them. The problems are just there in case you feel that you wish to clarify your intuition. --PST 09:51, 10 June 2009 (UTC)

but n.5 is not about connectedness.... ;-) --pma (talk) 16:59, 10 June 2009 (UTC)

but it does have something to do with disjoint non-empty open sets.... ;-) --PST 09:39, 12 June 2009 (UTC)

PST, will you give us some details about points 6 and 7? I was reflecting that a countable metric space (X,d) is always totally disconnected, which answers the OP about the connectedness of Q endowed with a p-adic metric. An argument is as follows. For simplifying the notations let's identify X with $\scriptstyle \mathbb {N}$. Fix ε>0, and consider the equivalence relation generated by the relation: nRm iff d(n,m)<2^-nε. Any equivalence class of this equivalence is open, and has diameter less than or equal to 4ε. Since ε>0 is arbitrary, this proves that the connected components of a point is the point itself. Question: what about more general topologies on $\scriptstyle \mathbb {N}$ e.g. Hausdorff? Thank you. --pma (talk) 19:16, 10 June 2009 (UTC)

Hmm -- it seems to me that a nonprincipal ultrafilter on N gives you a T1, but not Hausdorff, topology (by adjoining the empty set). I'm not sure whether there are any Hausdorff topologies on N other than the discrete topology — I'll go think about it on the exercise machine a little. --Trovatore (talk) 05:13, 11 June 2009 (UTC)

Well, there are many countable metric spaces, e.g. any countable subspace of R; in particular, any countable ordinal with its order topology. But are there non metrizable countable topological spaces that are T₂, or even more separated? If so, can they be connected? Can a countable topological space fail to be A₁? I think so, but do not have examples in mind at the moment...--pma (talk) 10:42, 11 June 2009 (UTC)

I think I have an elementary example:for each prime p and nonnegative n, let there be a basis set B-p-n of just those numbers containing p as a factor of multiplicity n. Defining the open sets to be the finite intersections and infinite unions of the basis sets seems to give a topology: it is Hausdorff since for any positive integers x,y there is a B-p-n that contains x but not y, and it is not discrete since every open set other than the empty set contains infinitely many points. — Charles Stewart (talk) 13:38, 11 June 2009 (UTC) PS I need to show more to show Hausdorff, although I think it is satisfied. — Charles Stewart (talk) 14:26, 11 June 2009 (UTC)

As PMajer pointed out, it is trivial to construct a nondiscrete countable space, one can take e.g. Q. The problem is to make it connected. Your space is totally disconnected. In fact, IIUIC it is homeomorphic to the subspace of the Baire space ω^ω consisting of functions which are zero in all but finitely many points. — Emil J. 14:15, 11 June 2009 (UTC)

My example is connected since (i) all pairs of points share open basis sets in common, and (ii) the only clopen sets are the empty set and Nat. — Charles Stewart (talk) 14:26, 11 June 2009 (UTC)

What is the exact definition of "B-p-n" then? If it consists of all numbers divisible by pⁿ but not pⁿ⁺¹, then every B-p-n is clopen, as its complement is the union of the sets B-p-m for m ≠ n, and the space is totally disconnected. If B-p-n consists of all numbers divisible by pⁿ, then the space is not Hausdorff, or even T₁, since every open set which includes a number x also contains all numbers divisible by x. — Emil J. 14:49, 11 June 2009 (UTC)

Ah! The former: thank you for leading me away from the path of misapprehension and error. And I see the resemblance to the chopped-off Baire topology, at last. — Charles Stewart (talk) 15:17, 11 June 2009 (UTC)

Separation axioms are not enough to make a countable space metrizable. Let F be any filter on ω which includes the Fréchet filter, and let X be the topological space with underlying set $\omega \cup \{\infty \}$ whose open sets are all subsets of ω, and sets of the form $U\cup \{\infty \}$, where $U\in F$. Note that any set containing $\infty$ is closed, and the closure of any $A\subseteq X$ is either A or $A\cup \{\infty \}$. X is Hausdorff, since {n} is clopen for all $n\neq \infty$. The space is in fact T₆: consider any closed set A. If $\infty \notin A$, then A is clopen, and a fortiori a zero set. If $\infty \in A$, define f: X → [0,1] by

$f(x)={\begin{cases}0,&x\in A,\\1/x&{\text{otherwise.}}\end{cases}}$

It is easy to see that f is continuous, and $f^{-1}(0)=A$. On the other hand, X is metrizable (if and) only if F has a countable filter basis. (For example, no ultrafilter F can have a countable basis.) The space is of course totally disconnected. — Emil J. 14:15, 11 June 2009 (UTC)

And what is A₁? — Emil J. 14:18, 11 June 2009 (UTC)

My solution to 6 and 7 is to take any totally ordered set (for example, $\scriptstyle \mathbb {N}$), and define a topology on it consisting of all upward closed sets. Any subspace is of the same form, and it's obviously hyperconnected. The space is T₀, but not T₁ (if it has more than one element). — Emil J. 11:12, 11 June 2009 (UTC)

As a matter of fact, if we generalize the setup to allow for total preorders (which give non-T₀ spaces), this is essentially the general form of spaces satisfying 7. Consider any such space. The condition implies that any 2-element subspace is either indiscrete or the Sierpiński space. If we define a preorder a ≤ b iff every open set containing a also contains b, then the latter condition implies that ≤ is total, and every open set is upward closed in ≤. Conversely, for every b > a, there is an open set which includes b but not a; since a union of open sets is open, every half-bounded interval (a, →) is an open set. Consequently, every upward closed set is open, except possibly for sets of the form [a, →) where a has no immediate predecessor. — Emil J. 12:32, 11 June 2009 (UTC)

Nice. In fact my curiosity was about: can a T2 countable space be connected, in particular non metrizable? Of course, connected and T1 are the mentioned hyperconnected cofinite set topology etc, but T2 connected seems more tricky. I'm not sure I understand Charles example: are you taking the collection of all B_n,p for all n and all primes p, or p is fixed? --pma (talk) 14:39, 11 June 2009 (UTC)

Here's another partial solution: every countable T₃ space is totally disconnected. Let X be any such space. For every n ≠ m, pick a pair of open sets separating n and m, and put them to B₀. For every finite intersection U of sets from B₀, and n ∈ U, put in B₁ a pair of disjoint open sets V and W such that n V and $X-U\subseteq W$. Construct B₂ from B₁ in the same way, and so on with B₃, B₄, …. Let B be the union of all B_k. By the construction, B is a countable basis of a T₃ topology Y. Such a topology is metrizable, hence Y is totally disconnected by your argument above. Since X is a finer topology than Y, X is totally disconnected, too. — Emil J. 15:26, 11 June 2009 (UTC)

And here's a simple direct proof for X T_3½. Let U be any neighbourhood of any point x. As X is T_3½, there exists a continuous function f: X → [0,1] such that f(x) = 0 and f(X − U) = 1. As X is countable, there exists a ∈ (0,1) − rng(f). Then f⁻¹([0,a)) is a clopen neighbourhood of x contained in U. It follows that X is 0-dimensional, and a fortiori, totally disconnected. — Emil J. 17:56, 11 June 2009 (UTC)

Hey, this thread is growing like the magic beanstalk! I will have a nice reading later. I was thinking that a good source of examples are inductive limits of countable spaces, like

$\scriptstyle X_{0}\subset X_{1}\subset ..X_{j}..\subset X_{\omega }:=\cup _{j\in \omega }X_{j}$.

So A is open in $\scriptstyle X_{\omega }$ iff for all j in $\omega$ $\scriptstyle A\cap X_{j}$ is open in $\scriptstyle X_{j}$. A nice example obtained this way is this: $\scriptstyle X:=(\mathbb {N} \times \mathbb {N} )\cup {\infty }$, where all points but $\scriptstyle \infty$ are open, and a base of neighborhoods of $\scriptstyle \infty$ are the epigraphs of functions $\scriptstyle \mathbb {N} \to \mathbb {N}$. So it is not A1, hence not metrizable. But it is T4: take two disjoint closed set: at least one of them does not contain $\scriptstyle \infty$, therefore it is open (its points are). I had some beer..is it ok? I see it is similar in fact homeomorphic to one of the previous ones by Emil; in fact T6 also holds. A1=first countable=points have countable basis of nbds. --pma (talk) 20:29, 11 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 B_ε(x, y) = {(x, y)} U N_ε(x + y/i) U N_ε(x - y/i) where N_ε(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. --PST 02:03, 12 June 2009 (UTC)

I would call it "irrational shadow topology" ;-) (the wall is Q×{0}). So, the irrationality of i ensures the T2 separation. You put that double shadow to give a non-empty boundary to any non-empty proper open set A. The key point is that the topology induced on the wall is the Euclidean... So, you first get a real number on the real boundary of A on the wall, then you approach it on the convenient side by rationals... I think I got it, maybe I need to check it. Very nice your example too! --pma (talk) 08:36, 12 June 2009 (UTC) Or also, one sees directly that a nonempty clopen is necessarily the whole space (it immediately invades everything).

I came up with a different example meanwhile. Let $\{\langle n_{k}^{0},n_{k}^{1}\rangle \mid k\in \omega \}$ be an enumeration of all pairs of distinct natural numbers, and let 2 < p₀ < p₁ < p₂ < … be a sequence of primes such that $n_{k}^{0}\not \equiv n_{k}^{1}{\pmod {p_{k}}}$. Put $B_{k}^{\alpha }=\{n\in \omega \mid n\equiv n_{k}^{\alpha }{\pmod {p_{k}}}\}$ for k ∈ ω and α = 0, 1. Let X be the topological space with underlying set ω and subbasis $\{B_{k}^{\alpha }\mid k\in \omega ,\alpha <2\}$. Note that X has a basis consisting of sets of the form $B_{k_{1}}^{\alpha _{1}}\cap \dots \cap B_{k_{r}}^{\alpha _{r}}$ where k₁ < k₂ < … < k_r, and each such set is nonempty by the Chinese remainder theorem. The closure of $B_{k}^{\alpha }$ is $X-B_{k}^{1-\alpha }$. 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)

Impressive! Thanks to you both. So, these are T2 spaces with countable basis, yet non metrizable. It seems that for countable spaces, T2 is still quite a wild class, right? Maybe countable metric have a decent classification? For instance, a countable metric space with no isolated points is homeomeorphic to Q (a theorem of Sierpinski). Or maybe there is a reasonable classification for compact countable spaces? I think I saw a number of papers about countable topological spaces, but I wonder if there is a foundational treatment. --pma (talk) 11:06, 13 June 2009 (UTC)

As for countable compact spaces: their iterated Cantor–Bendixson derivative eventually becomes empty, which gives you a sort of classification. — Emil J. 11:08, 15 June 2009 (UTC)