User henno brandsma - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-24T22:31:42Z http://mathoverflow.net/feeds/user/2060 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/122528/a-question-on-metrizable-space/122578#122578 Answer by Henno Brandsma for A question on metrizable space Henno Brandsma 2013-02-21T21:54:06Z 2013-02-21T21:54:06Z <p>For a metric space $(X,d)$ and an infinite cardinal number $\kappa$, the following are equivalent:</p> <ol> <li>$X$ has a base of cardinality $\le \kappa$.</li> <li>X has a network of cardinality $\le \kappa$. (A network is a collection $\mathcal{N}$ of subsets of $X$ such that every open set is a union of elements from $\mathcal{N}$; a base is just a network that consists of open sets.)</li> <li>Every open cover of $X$ has a subcover of cardinality $\le \kappa$.</li> <li>Every closed discrete subspace $A$ of $X$ has cardinality $\le \kappa$.</li> <li>Every discrete subspace $A$ of $X$ has cardinality $\le \kappa$.</li> <li>Every pairwise disjoint family of non-empty open sets of $X$ has cardinality $\le \kappa$.</li> <li>$X$ has a dense subspace of cardinality $\le \kappa$.</li> </ol> <p>$1)\rightarrow 2)$ is obvious, and true for all topological spaces $X$.</p> <p>$2)\rightarrow 3)$ is true in general as well: Let $\mathcal{N}$ be a network with $\left|\mathcal{N}\right| \le \kappa$. If $\mathcal{U} = \left\{ U_i : i \in I \right\}$ is an open cover of $X$, then for each $x \in X$ we pick $i(x) \in I$ and $N_x \in \mathcal{N}$, such that $x \in N_x \subset U_{i(x)}$. Then $\left\{N_x : x \in X\right\} = \mathcal{N}'$ has cardinality $\le \kappa$, and for each distinct element $A$ from $\mathcal{N}'$ we pick $U(A)$ from $\mathcal{U}$ with $A \subset U(A)$ ($A = N_x$ for some $x$, and we pick $U(A) = U_{i(x)}$). Then $\left\{U(A) : A \in \mathcal{N}'\right\}$ is the required subcover.</p> <p>$3)\rightarrow 4)$ is always true as well: Let $A$ be closed and discrete. Each $x \in A$ has an open neighbourhood $U_x$ that intersects $A$ in $\{x\}$ only. The open cover $\mathcal{U} = \left\{U_x : x \in A\right\} \cup \{X \setminus A\}$ cannot spare any $U_x$ (or $x$ will not be covered), so the cover $\mathcal{U}$ has cardinality $|A|$ and no subcover of cardinality strictly less than $|A|$. So $|A| \le \kappa$, or we'd have a contradiction with 3).</p> <p>$4)\rightarrow 5)$ Here we need only perfect normality of $X$, in the sense only that each open set is a countable union of closed sets, or equivalently that each closed set is a $G_\delta$. Let $A$ be discrete, then I claim that $A$ is open in $\overline{A}$. </p> <p>Proof of claim (needs only that singletons are closed): let $x$ be in $A$ and let $U_x$ be an open neighbourhood of $x$ that intersects $A$ only in $\{x\}$. This $U_x$ has the property that $\overline{A} \cap U_x = \{x\}$ as well: $y \neq x$ and $y \in \overline{A} \cap U_x$, then $U_x\setminus\{x\}$ is an open neighbourhood of $y$, $y \in \overline{A}$ so $U_x\setminus\{x\}$ must intersect $A$, but this can only happen in $\{x\}$, contradiction, so that $\{x\}$ is open in $\overline{A}$. </p> <p>But then, as $A$ is perfectly normal (being metrisable), $A = \cup_{i \in \mathbb{N}} A_i$ where the $A_i$ are closed in $\overline{A}$ (and thus closed in $X)$. So the $A_i$ are closed and discrete, and by 4) we have $|A_i| \le \kappa$. So $|A| \le \aleph_0 \cdot \kappa = \kappa$, as well.</p> <p>$5)\rightarrow 6)$ is true for all topological spaces: pick $x_i \in U_i$ for any pairwise disjoint family $\left\{U_i : i \in I\right\}$ of non-empty open sets. By definition we have that $\left\{x_i: i \in I\right\}$ is discrete (as witnessed by the $U_i$), and so $\left|I\right| \le \kappa$, and 6) has been proved.</p> <p>$6)\rightarrow 7)$ Here we need the metric in a more essential way. For each $n \in \mathbb{N}$, let $D_n$ be a family of points with the property that $x,y \in D_n$ with $x \neq y$ implies $d(x,y) \ge \frac{1}{n}$, and $D_n$ is maximal with that property. Here we use Zorn's lemma, or some equivalent principle. Note that the balls with radius $\frac{1}{2n}$ around the points of $D_n$ are disjoint so that $|D_n| \le \kappa$ by 6).</p> <p>Let $D = \cup_n D_n$, we claim that $D$ is dense in $X$. We already see that $D$ is of the right size, as $|D| \le \aleph_0 \cdot \kappa = \kappa$. For if $x$ is not in $\overline{D}$, we have that $d(x,\overline{D}) > 0$ and so for some $m \in \mathbb{N}$ we know that $d(x,\overline{D}) > \frac{1}{m}$. But then, for this $m$, $d(x,\overline{D_m}) \ge d(x,\overline{D}) > \frac{1}{m}$ and in particular: $d(x,y) > \frac{1}{m}$ for all $y \in D_m$. But then we could have added $x$ to $D_m$ and would have obtained a strictly larger $D_m$, and this cannot be. So $D$ is dense.</p> <p>$7)\rightarrow 1)$ This needs the metric "most". Let $D$ be the dense subset of cardinality at most $\kappa$. Let $\mathcal{B} = \left\{B(x,r): x \in D; r \in \mathbb{Q}\right\}$, then $\left|\mathcal{B}\right| \le \aleph_0 \cdot \kappa = \kappa$. I claim that $\mathcal{B}$ is a base for $X$: let $U$ be open and $x \in U$. Some $\epsilon>0$ exists such that $B(x,e) \subset U$, and as $D$ is dense there is some $y \in D$ in $B(x,\frac{\epsilon}{3})$. Now pick $r \in \mathbb{Q}$ such that $\frac{\epsilon}{3} &lt; r &lt; \frac{\epsilon}{2}$, then $x \in B(y,r)$ (which is from $\mathcal{B}$) and $B(y,r) \subset B(x,\epsilon)$: if for some $z$, $d(z,y) &lt; r$ then $d(z,x) \le d(z,y) + d(y,x) &lt; r + r &lt; \epsilon$, and so there is a $B_x = B(y,r)$ from $\mathcal{B}$ such that $x \in B_x \subset U$, as required for a base.</p> <p>This concludes the proof of the equivalence, which shows that weight, network weight, Lindelöf number, extent, cellularity and other cardinal invariants are all the same for metrisable spaces. </p> http://mathoverflow.net/questions/19930/writing-papers-in-pre-latex-era/19953#19953 Answer by Henno Brandsma for Writing papers in pre-LaTeX era? Henno Brandsma 2010-03-31T14:46:43Z 2012-12-02T04:59:14Z <p>On my university some people used troff (<a href="http://en.wikipedia.org/wiki/Troff" rel="nofollow">http://en.wikipedia.org/wiki/Troff</a>) for that. At least until the early nineties. There are all sorts of math macros in that too. It's not as nice looking as $\TeX$, but it does the job. We used Solaris, so it came with that program. I myself have only used $\TeX$ but I have typeset old stuff in troff.</p> http://mathoverflow.net/questions/105117/countable-topological-spaces-of-uncountable-weight/109631#109631 Answer by Henno Brandsma for countable topological spaces of uncountable weight Henno Brandsma 2012-10-14T17:56:22Z 2012-10-14T17:56:22Z <p>Another favourite example of mine: the Hewitt-Marczewski-Pondiczery theorem says that the product of continuum or fewer separable spaces is separable. So $\mathbb{R}^\mathbb{R}$ has a countable dense subset $D$ and it is clear any countable regular space is normal (from being Lindelöf), and it's not hard to show that every point of $D$ has a local base of $\mathfrak{c}$ many open sets (and not fewer). So it's not like the other examples mentioned, in that there are no isolated points.</p> http://mathoverflow.net/questions/84950/metrizable-space/84995#84995 Answer by Henno Brandsma for metrizable space Henno Brandsma 2012-01-05T19:25:37Z 2012-01-05T19:25:37Z <p>No to all your questions. There are lots of countable non-metrizable spaces: an easy one is $\mathbb{N}$ in the cofinite topology, which can be written as a countable (disjoint) union of singletons (which are metrizable, and closed and compact). Using ultrafilter spaces (given an ultrafilter $\mathcal{F}$ on $\mathbb{N}$, define $X = \mathbb{N} \cup {\infty}$ where $\mathbb{N}$ is discrete and a neighbourhood of $\infty$ is of the form $A \cup {\infty}$, where $A \in \mathcal{F}$); such spaces are countable, hereditarily normal but not metrizable (not even first countable at $\infty$). </p> <p>All these spaces above are $\sigma$-compact but not compact, and of course a discrete countable set like $\mathbb{N}$ is metrizable, $\sigma$-compact but not compact, as is $\mathbb{R}$, e.g. </p> <p>Also $[0,1]$ is pseudocompact, but the $\sigma$-compact subset $\{ \frac{1}{n} \mid n \in \mathbb{N} \}$ is not (pseudo)-compact.</p> http://mathoverflow.net/questions/84726/some-questions-on-lindelof-property/84905#84905 Answer by Henno Brandsma for some questions on Lindelöf property Henno Brandsma 2012-01-04T21:45:03Z 2012-01-04T21:45:03Z <p>A space is Lindelöf iff it satisfies condition A and is metaLindelöf (every open cover has a point-countable refinement), so one could say the difference is metaLindelöfness.</p> http://mathoverflow.net/questions/84703/does-every-lindelof-uniform-space-have-a-lindelof-completion/84770#84770 Answer by Henno Brandsma for Does every Lindelof uniform space have a Lindelof completion? Henno Brandsma 2012-01-02T22:15:16Z 2012-01-03T06:55:40Z <p><a href="http://projecteuclid.org/DPubS/Repository/1.0/Disseminate?view=body&amp;id=pdf_1&amp;handle=euclid.pjm/1102970055" rel="nofollow">this</a> paper by Howes gives a characterization of uniform spaces with a Lindelöf completion, but the characterization uses the derived uniformity, and notions like preparacompactness. I don't see as yet how it would imply, or refute, your question, but the link might help anyway. Howes did a write up of all the theory of uniform spaces and covering properties in his book "Modern Analysis and Topology" as well, and the result can also be found in its 4th chapter.</p> <p>[added] It is equivalent for a Lindelöf uniform space $X$ to have a paracompact or a Lindelöf completion $X^{\ast}$, by the classical theorem (5.1.25 in Engelking, General Topology): a paracompact space that has a Lindelöf dense subspace is Lindelöf, and $X$ is dense in $X^{\ast}$.</p> http://mathoverflow.net/questions/79435/question-about-0-dimensional-polish-spaces/79454#79454 Answer by Henno Brandsma for Question about 0-dimensional Polish spaces Henno Brandsma 2011-10-29T08:09:35Z 2011-10-29T08:09:35Z <p>If you have such a union of $O = \cup_n B_n$, consider the sets $B'_n = B_n \setminus \cup_{i=0}^{n-1} B_i$, where $B'_0 = B_0$. Show these sets are disjoint, clopen, and have the same union as the original sets, as for every $x \in O$ there is a first index $n(x)$ such that $x \in B_{n(x)}$. </p> http://mathoverflow.net/questions/78975/ccc-collectionwise-normality-paracompact/79021#79021 Answer by Henno Brandsma for CCC +　collectionwise normality =>　paracompact? Henno Brandsma 2011-10-24T20:49:03Z 2011-10-24T20:57:52Z <p>Yes, there is. Let $I = \omega_1$ be the first uncountable ordinal, and let $P = \{0,1\}^I$ be the uncountable product of discrete spaces of 2 points. Let $S$, the so-called $\Sigma$-product be its subspace of all points that have at most countably many coordinates different from $0$. </p> <p>It is well known that $S$ is ccc (as a dense subset of a ccc space $P$) and countably compact (but not compact, being dense in $P$) and (hereditarily) collectionwise normal, but not paracompact (being countably compact and non-compact). Proofs of some of these facts can be found <a href="http://dantopology.wordpress.com/2009/12/12/a-note-about-sigma-product-of-compact-spaces/" rel="nofollow">here</a>, e.g.</p> <p>Corson showed in <a href="http://www.jstor.org/pss/2372929" rel="nofollow">this paper</a> (cannot find free download) that if $X$ is dense in a product of metrizable spaces, and $X \times X$ is normal, then $X$ is collection wise normal. This can be used to show the collectionwise normality, as $S \times S$ is homeomorphic to $S$, so one only needs to show normality. </p> <p>A very related example is the set $C_p(L(\aleph_1))$, where $L(\aleph_1)$ is the one-point compactification of a discrete space of size $\aleph_1$, and $C_p(X)$ is the space of continuous real-valued functions on a space $X$, in the subspace topology of $\mathbb{R}^X$. This example is discussed on page 113 of the book General Topology III, in the Encyclopedia of Mathematical Sciences series (volume 51). All spaces $C_p(X)$ are ccc, and if they are normal, they are collectionwise normal (due to Reznichenko), so it's natural to look for examples there.</p> <p>As an aside: by well-known results, both these spaces are Fréchet-Urysohn, but not first countable. Can there be first countable examples ?</p> http://mathoverflow.net/questions/78196/on-generalized-ordered-spaces/78207#78207 Answer by Henno Brandsma for On generalized ordered spaces Henno Brandsma 2011-10-15T13:00:12Z 2011-10-15T13:00:12Z <p>Lemma: if $K$ and $L$ are (order) convex in a linearly ordered set $X$, and $x$ is in $K \cap L$, then $K \cup L$ is convex as well.</p> <p>Proof: suppose $a &lt; b$ are in $K \cup L$ and $c$ lies in $(a,b)$. Consider 2 cases:</p> <p>1) $c \le x$: then $a$ is in one of the convex sets, say $K$, and so is $x$, but then $x \in (a, x]$ is in $K$ as well, and we are done.</p> <p>2) $x &lt; c$: then we have the same argument with $x$ and $b$.</p> <p>Now, if $X$ is a GO-space, $G$ an open subset of $X$, and $K$ a maximal convex subset of $G$, then let $x$ be an arbitrary point of $K$. As $X$ is a GO-space, $x$ has a local base of convex neighbourhoods, so pick a convex neighbourhood $U$ of $x$ such that $U \subset G$. By the lemma, as $K$ and $U$ intersect (in $x$) and are both convex, their union is a convex subset of $G$ and by maximality of $K$, $K \cup U \subset K$, or $U \subset K$, showing $x$ is an interior point of $K$; hence $K$ is open.</p> <p>Note that the proof is completely analogous to the proof of the fact that in a locally connected space the components of the open sets are open; there we use that the union of intersecting connected sets is connected and the fact that all points have local bases of connected neighbourhoods, here we use the same with connected replaced by convex.</p> http://mathoverflow.net/questions/77905/topology-generated-by-the-collection-of-open-sets/77909#77909 Answer by Henno Brandsma for Topology generated by the collection of open sets Henno Brandsma 2011-10-12T09:21:44Z 2011-10-12T09:21:44Z <p>So we start with $(X, \mathcal{T})$, a $T_1$ space in which every point is a $G_{\delta}$, as witnessed by open sets $U_n(x)$, $n \in \mathbb{N}$, $x \in X$. W.l.o.g. we can take these sets to be decreasing.</p> <p>The topology generated by all these sets we call $\mathcal{T}'$, say, and it is $T_1$, because for every $x \neq y$, there is some $U_n(x)$ that does not contain $y$ (or else $y$ would be in their intersection, for all $n$, and this intersection is precisely $\{x\}$), and this witnesses the $T_1$ property ($\{y\}$ is closed, by this argument).</p> <p>A base for this topology is given by all finite intersections of the sets $U_n(x)$. It's not clear how this topology would be first countable; the obvious candidate for a local base at $x$ is all sets $U_n(x)$, but if $z$ is in $U_n(x) \cap U_m(y)$, I don't see how we automatically get some $U_m(z)$ inside it. So in general, this might not work as advertised. </p> <p>This might be remedied by using elementary submodels (see e.g. <a href="http://www.math.uncc.edu/~adow/Ftp/Intro.Elem.Subm/elem.ps" rel="nofollow">http://www.math.uncc.edu/~adow/Ftp/Intro.Elem.Subm/elem.ps</a>) I think. This would also then do the Hausdorff case, which this construction (even if it could be made to work) does not always give, e.g. if we take a countable nowhere first countable space and use the cofinite topology on it. This is a $T_1$ condensation obtained by the described construction, which is not Hausdorff.</p> http://mathoverflow.net/questions/66759/collatz-conjecture-solved Collatz conjecture solved? Henno Brandsma 2011-06-02T18:03:03Z 2011-06-02T18:03:03Z <p>In <a href="http://preprint.math.uni-hamburg.de/public/papers/hbam/hbam2011-09.pdf" rel="nofollow">this paper</a> Gerhard Opfer claims to have solved Collatz conjecture. Looks like a strange paper to me, but I haven't studied it. How legit does it look to people more specialized in these matters? It would be cool to see it solved....</p> http://mathoverflow.net/questions/66650/connectifications/66652#66652 Answer by Henno Brandsma for Connectifications? Henno Brandsma 2011-06-01T12:05:19Z 2011-06-01T15:20:01Z <p>They do not always exist (I believe the Sorgenfrey line does not have one, e.g.), and if they exist they are not very well-behaved. </p> <p><a href="http://www.auburn.edu/~gruengf/papers/connect.pdf" rel="nofollow">This might be a relevant paper</a></p> http://mathoverflow.net/questions/66240/topological-spaces-uncountable-subsets-and-separability/66271#66271 Answer by Henno Brandsma for Topological spaces, uncountable subsets and separability Henno Brandsma 2011-05-28T11:37:27Z 2011-05-28T11:37:27Z <p>The property "Every uncountable set has a limit point" is related to the Lindelöf property (every open cover has a countable subcover). For metrisable spaces these notions are equivalent, and in general Lindelöf implies the limit point property. And there are many Lindelöf spaces that are not separable, as the other examples show. For metrisable spaces, being Lindelöf, separable, second countable, etc. are all equivalent. </p> http://mathoverflow.net/questions/65811/characterisation-of-paracompact-spaces-by-some-sort-of-embeddability/65888#65888 Answer by Henno Brandsma for Characterisation of paracompact spaces by some sort of embeddability? Henno Brandsma 2011-05-24T18:46:38Z 2011-05-24T18:46:38Z <p>You cannot embed all paracompact spaces in this way. Even if we allow at most countably many non-zero coordinates: in that case we get subspaces of $\Sigma$-products, which have been well-studied in general topology and functional analysis. Compact subspaces of these are called Corson compact spaces, and none of these map onto $[0,1]^{\omega_1}$. So the latter space is an example of a compact Hausdorff space (so paracompact etc.) that cannot be embedded into a $\Sigma$-product of copies of $I$ (or $R$).</p> http://mathoverflow.net/questions/60457/elementaryshortuseful/60657#60657 Answer by Henno Brandsma for Elementary+Short+Useful Henno Brandsma 2011-04-05T07:52:26Z 2011-04-05T07:52:26Z <p>The well-ordering theorem and an application (that uses transfinite recursion, after well-ordering a set). Many interesting sets and examples can be built that way. Or maybe Axiom of Choice/Zorn's lemma (show one from the other) and then show the well-ordering theorem.</p> http://mathoverflow.net/questions/60375/is-r3-the-square-of-some-topological-space/60389#60389 Answer by Henno Brandsma for Is R^3 the square of some topological space? Henno Brandsma 2011-04-02T21:25:00Z 2011-04-02T21:25:00Z <p><a href="http://blog.plover.com/math/R3-root.html" rel="nofollow">this blog post</a> refers to some papers with proofs. I've heard Robert Fokkink explain his proof and there he also told us the cohomological proof, which generalizes it to all Euclidean spaces of odd dimension.</p> http://mathoverflow.net/questions/56939/must-a-linearly-ordered-separable-space-be-metrizable/57022#57022 Answer by Henno Brandsma for Must a linearly ordered, separable space be metrizable? Henno Brandsma 2011-03-01T18:39:53Z 2011-03-01T18:39:53Z <p>You already found a (classical) counterexample: the double arrow ($[0,1] \times \{0,1\}$, ordered lexicographically), which is even compact and separable. There is however a nice metrization theorem for linearly ordered spaces (due to <a href="http://www.ams.org/journals/proc/1969-022-02/S0002-9939-1969-0248761-1/S0002-9939-1969-0248761-1.pdf" rel="nofollow">Lutzer</a>): a linearly ordered space $X$ is metrizable (in the order topology) iff the diagonal $D = \{(x,x) : x \in X\}$ is a countable intersection of open subsets of $X \times X$ (a $G_\delta$). This condition is also necessary and sufficient for countably compact regular spaces as well, not just the ordered ones.</p> http://mathoverflow.net/questions/35408/naturally-occuring-groups-with-cardinality-greater-than-the-reals/57020#57020 Answer by Henno Brandsma for Naturally occuring groups with cardinality greater than the reals. Henno Brandsma 2011-03-01T18:26:51Z 2011-03-01T18:26:51Z <p>Any product group like $\{0, 1\}^I$ for index sets $I$, using mod 2 addition coordinatewise. It's just (isomorphic to) the power set of $I$ using symmetric difference as the addition. It's of course also a ring (pointwise multiplication / intersection ). These Boolean groups often come up in general topology.</p> http://mathoverflow.net/questions/53300/locally-compact-hausdorff-space-that-is-not-normal/56805#56805 Answer by Henno Brandsma for Locally compact Hausdorff space that is not normal Henno Brandsma 2011-02-27T10:07:38Z 2011-02-27T10:14:51Z <p>Another nice elementary example is the rational sequence topology. For every irrational number $x$ we pick a sequence $q(x)_n$ of rational numbers, all different, that converge to $x$ (in the usual topology on $\mathbb{R}$). A topology on $\mathbb{R}$ is then defined by specifying basic neighbourhoods: a rational number $q$ has $\{ q \}$ as a basic neighbourhood (it is isolated), while an irrational number $x$ has basic neighbourhoods of the form $\{ q(x)_n : n \ge k \}$, $k \in \mathbb{N}$. One checks that this defines a topology in which the irrationals are closed and discrete (in itself), $\mathbb{Q}$ is dense (and open), and $X$ is Hausdorff and zero-dimensional (basic open sets are clopen, this uses that 2 sequences that converge to 2 different irrationals only have at most finitely many terms in common, so no basic neighbourhood can have another irrational in its closure), so Tychonov, and locally compact, as all basic neighbourhoods are compact (finite or convergent sequences). But by Jones' lemma (in a normal space, with dense set $D$ and closed discrete set $A$ we have that $2^{|A|} \le 2^{|D|}$), a proof of which can be found <a href="http://at.yorku.ca/cgi-bin/bbqa?forum=ask_a_topologist_2006;task=show_msg;msg=0578.0001.0001.0001.0001" rel="nofollow">here, e.g.</a>) we have that this space is not normal. </p> http://mathoverflow.net/questions/29624/how-many-orders-of-infinity-are-there/29625#29625 Answer by Henno Brandsma for How many orders of infinity are there? Henno Brandsma 2010-06-26T17:59:46Z 2011-02-23T19:53:28Z <p>Yes, this is possible, if you define the order to be dominance with finitely many exceptions. So f &lt; g iff the set of n with f(n) > g(n) is finite. What you call a complete system of growth functions is called a dominating subset of $\omega^\omega$ (and a scale if it is well-ordered). See van Douwen's paper "The integers and topology" in the Handbook of Set Theoretic Topology. The minimal cardinality of such a dominating family is called $\mathfrak{d}$ in the set-theoretic literature and it's one of the so-called cardinal invariants of the continuum. What is known is that its cofinality is at least $\mathfrak{b}$ where the latter is the minimal size of an unbounded set in $\omega^\omega$ in the partial order of eventual dominance. Also, $\mathfrak{d}$ is equal to the minimal size of a cofinal subset of $\omega^\omega$ in the total dominance order that you defined. So indeed, the problem is the same for both orders, and both have minimal size $\mathfrak{d}$. The eventual dominance is more commonly used though, and that's how I knew it at first.</p> <p>This cardinal can assume almost any value (under said restriction on the cofinality at least) and there has been a lot of study on this and similar cardinal invariants and their interrelations. We can have $\omega_1 = \mathfrak{d} &lt; \mathfrak{c}$, $\omega_1 &lt; \mathfrak{d} &lt; \mathfrak{c}$ and $\omega_1 &lt; \mathfrak{d} = \mathfrak{c}$, in different models of ZFC.</p> http://mathoverflow.net/questions/56275/is-a-connected-separable-locally-euclidean-hausdorff-topological-space-second-cou/56290#56290 Answer by Henno Brandsma for Is a connected separable locally euclidean Hausdorff topological space second countable? Henno Brandsma 2011-02-22T15:26:28Z 2011-02-22T15:26:28Z <p>Indeed, check <a href="http://arxiv.org/pdf/0910.0885v1" rel="nofollow">the paper by Gauld</a>. Your (4) implies his condition hemicompact. His example at p15, that you saw refutes the just separable condition (7). Note that (1)-(6) imply imply metrisability for just continuous manifolds, so it still might be that the situation vis à vis separability is different for <strong>smooth</strong> manifolds instead of continuous ones, though I suspect not.</p> http://mathoverflow.net/questions/43069/is-there-a-list-of-all-connected-t-0-spaces-with-5-points/43077#43077 Answer by Henno Brandsma for Is there a list of all connected T_0-spaces with 5 points? Henno Brandsma 2010-10-21T17:27:34Z 2010-10-21T17:27:34Z <p>At the <a href="http://www.research.att.com/~njas/sequences/index.html" rel="nofollow">online encyclopedia of integer sequences</a> we find, when we type T_0 topologies several hits. Sequence A028856 is the sequence of homeomorphism classes of T_0 topologies, and A028858 has all connected ones (308 topologies of which 235 connected, on 5 points). No explicit list of spaces, though, but some literature references that might help.</p> http://mathoverflow.net/questions/42117/discrete-subspaces-of-hausdorff-spaces/42385#42385 Answer by Henno Brandsma for Discrete subspaces of Hausdorff spaces Henno Brandsma 2010-10-16T13:01:14Z 2010-10-16T13:01:14Z <p>In a more general light:</p> <p>folklore theorem: Every infinite topological space contains a homeomorphic copy of one (or more) of the following 5 spaces:</p> <ol> <li>$\mathbf{N}$ in the indiscrete topology (only $\mathbf{N}$ and $\emptyset$ are open).</li> <li>$\mathbf{N}$ in the co-finite topology (only $\mathbf{N}$ and all finite sets are closed).</li> <li>$\mathbf{N}$ in the upper topology (the empty set and all sets $U(k) = \{ n \in \mathbf{N} : n \ge k \}$, $k \in \mathbf{N}$, are open).</li> <li>$\mathbf{N}$ in the lower topology ($\mathbf{N}$, $\emptyset$, and all sets $L(k) = \{ n \in \mathbf{N} : n \le k \}$, $k \in \mathbf{N}$, are open).</li> <li>$\mathbf{N}$ in the discrete topology (all subsets are open).</li> </ol> <p>As each of the spaces has the property that every infinite subspace of it is homeomorphic to the whole space, this list is minimal.</p> <p>And spaces 1-4 are not Hausdorff, which implies what you need, as being Hausdorff is hereditary.</p> <p>The nicest proof of this I know uses Ramsey's theorem (off hand I do not know a reference, who does?) using a partition of the triples or pairs of X, IIRC. </p> http://mathoverflow.net/questions/9667/what-are-some-results-in-mathematics-that-have-snappy-proofs-using-model-theory/35190#35190 Answer by Henno Brandsma for What are some results in mathematics that have snappy proofs using model theory? Henno Brandsma 2010-08-11T04:08:06Z 2010-08-11T04:08:06Z <p>Alan Dow and others have explored the use of elementary submodels in Topology. See e.g.his introductory paper <a href="http://math.uncc.edu/~adow/Ftp/Intro.Elem.Subm/elem.ps" rel="nofollow">here</a>. One application: the theorem by Arhangel'lskij that a Hausdorff Lindelöf first countable space is at most size continuum. There is a technical proof using transfinite recursion (the standard one), but also a slick one using elementary submodels (of sufficiently large countable models of ZFC). </p> http://mathoverflow.net/questions/30661/non-homeomorphic-spaces-that-have-continuous-bijections-between-them Non-homeomorphic spaces that have continuous bijections between them Henno Brandsma 2010-07-05T20:03:58Z 2010-08-09T18:04:27Z <p>What are nice examples of topological spaces $X$ and $Y$ such that $X$ and $Y$ are <strong>not</strong> homeomorphic but there do exist continuous bijections $f: X \mapsto Y$ and $g: Y \mapsto X$?</p> http://mathoverflow.net/questions/33947/topological-spaces-that-resemble-the-space-of-irrationals/33982#33982 Answer by Henno Brandsma for Topological spaces that resemble the space of irrationals Henno Brandsma 2010-07-31T06:17:02Z 2010-07-31T06:17:02Z <p>As regards Q (your first remark), it is true that all countable metrisable spaces without isolated points are homeomorphic to Q. If you want to omit metrisable, replace it by T_3 and second countable. One then notes that a dense subset of R^n doesn't have isolated points, and is metrisable.</p> http://mathoverflow.net/questions/32788/lebesgue-dimension-of-images/32813#32813 Answer by Henno Brandsma for Lebesgue dimension of images Henno Brandsma 2010-07-21T16:40:05Z 2010-07-21T16:40:05Z <p>Some results from Engelking's dimension theory book:</p> <p>If $f: X \mapsto Y$ is a closed, continuous and surjective function between normal spaces $X$ and $Y$, and $\forall y \in Y: | f^{-1}[{y}] | \le k$ for some integer $k \ge 1$, then $\dim(Y) \le \dim(X) + (k-1)$. </p> <p>If $f: X \mapsto Y$ is an open, continuous and surjective function between a weakly paracompact space $X$ and a normal space $Y$ with finite fibres, then $\dim(X) = \dim(Y)$.</p> <p>If $f: X \mapsto Y$ is an open, continuous and surjective function between a locally compact normal space $X$ and a weakly paracompact normal space $Y$ such that all fibres are at most countable, then $\dim(Y) \le \dim(X)$.</p> <p>If $f: X \mapsto Y$ is an open-and-closed, continuous and surjective function between a locally compact normal space $X$ and a paracompact space $Y$ such that all fibres are at most countable, then $\dim(Y) \le \dim(X)$.</p> <p>In the other direction: If $f: X \mapsto Y$ is a closed, continuous function between a normal space $X$ and a paracompact space $Y$ such that the dimension of the fibres is at most 0 (this includes the empty fibres of dimension -1), then $\dim(X) \le \dim(Y)$.</p> <p>weakly paracompact includes Hausdorff, and is called metacompact or point-paracompact by other authors: every open cover has a point-finite refinement.</p> http://mathoverflow.net/questions/32035/are-nets-and-filters-useful-in-geometry-and-topology/32043#32043 Answer by Henno Brandsma for Are nets and filters useful in geometry and topology? Henno Brandsma 2010-07-15T17:06:01Z 2010-07-15T17:06:01Z <p>I think the net formulation is quite useful to know, at least. Analysts seem to be most fond of it, as they naturally work with sequences anyway. E.g. on easily shows that the closure of a subgroup $H$ in a topological group $G$ is a subgroup: just note that for $x,y$ in closure of $H$, we find nets (wlog with the same index set) $(x_i), (y_i)$ from $H$ that converge to $x$, resp. $y$, and then $x_i \cdot {y_i}^{-1} \rightarrow x \cdot y^{-1}$ from continuity of the group operations, and the left hand side lives in $H$, so the right hand side is in closure H, and this is thus a subgroup. </p> <p>Similarly, $A \cdot B$ is closed in $G$ when $A$ is closed and $B$ is compact: take a net $(x_i \cdot y_i)_{i \in I}$ in $A \cdot B$ converging to $z$. By compactness the net $(y_i)$ has a subnet converging to $y \in B$, indexed by $J$ say, and then the corresponding subnet $x_j = x_j \cdot y_j \cdot y_j^{-1}$ converges to $z \cdot y^{-1}$ and as the $(x_j)$ are in $A$, so is the limit $z \cdot y^{-1}$ and so $z = (z \cdot y^{-1} \cdot y)$ is in $A \cdot B$, making it closed.</p> <p>Nets also make for a nice formulation of the Riemann integral (using partitions on the interval as a directed set under refinement) as a limit of a certain net. Some proofs just look more "natural" in a net formulation, I think. One can do the proof for sequences first, and see how it generalizes using nets. </p> http://mathoverflow.net/questions/31600/question-about-closed-projection/31957#31957 Answer by Henno Brandsma for Question about closed projection Henno Brandsma 2010-07-15T04:22:56Z 2010-07-15T06:16:21Z <p>Let $(y_n)$ be a sequence in $Y$. Let $A$ be the subset of $Y \times \mathbf{R}$ of all points $(y_n, \frac{1}{n})$ for $n \in \mathbf{N}$, and let $B$ be its closure. Then $\pi_2[B]$ is closed in $\mathbf{R}$, and contains all points $\frac{1}{n}$, so it contains $0$. So for some $y \in Y$, $(y,0) \in B$. Using the countable base we can extract a subsequence of the $(y_n)$ that converges to $Y$ (as $Y$ is first countable in particular). We do then need that $y$ is in the closure of all subsequences of $(y_n)$ as well, which follows in a similar way, otherwise we cannot get (without separation axioms) a convergent subsequence from first countability alone. But this works.</p> <p>So $Y$ is sequentially compact, which implies that $Y$ is countably compact (in the covering sense; no separation axioms needed) and as $Y$ is also Lindelöf, being second countable, $Y$ is compact. </p> http://mathoverflow.net/questions/30662/relatively-countably-compact-subsets-without-countably-compact-closure Relatively countably compact subsets without countably compact closure. Henno Brandsma 2010-07-05T20:09:48Z 2010-07-08T09:41:11Z <p>I am looking for a (Hausdorff or better) space $X$ and a subset $A$ of $X$ that is relatively countably compact (every sequence from $A$ has an accumulation point in $X$) such that the closure of $A$ is not countably compact. It is known that in many "nice" spaces such examples do not exist (a classical case being normed spaces in their weak topology).</p> <p>Edit: for $T_4$ spaces this cannot happen, as the closure of a relatively countably compact subset is pseudocompact (Suppose A is relatively countably compact. If f from cl(A) to R is unbounded then f|A is unbounded as well, as A is dense in cl(A). So we can find {x_n: n in N} in A such that |f(x)| >= n. Let p in cl(A) be an accumulation point of this set, by A being relatively countably compact. Then continuity at f implies that |f(x_n)| &lt;= |f(p)| + 1, for all but finitely many n. This contradicts the choice of the x_n, contradiction.) So cl(A) is pseudocompact, and hence in a normal space, countably compact. This explains the properties of the example given below. </p> http://mathoverflow.net/questions/131206/whats-the-definition-of-continuous-of-set-valued-functions/131209#131209 Comment by Henno Brandsma Henno Brandsma 2013-05-22T04:04:33Z 2013-05-22T04:04:33Z Also, this Vietoris topology is the one induced by the Hausdorff metric, in the special case where $X$ is a compact metric space and we take only the closed non-empty subsets (instead of the whole power set). The closed non-empty subsets of a space $X$ in this topology are called the hyperspace of $X$, denoted $H(X)$ or sometimes $2^X$. http://mathoverflow.net/questions/122528/a-question-on-metrizable-space/122578#122578 Comment by Henno Brandsma Henno Brandsma 2013-02-25T19:02:42Z 2013-02-25T19:02:42Z The term was probably in Russian first; it was indeed Arhangel'skij that first introduced it. It's called a network in Engelking, e.g., and in many papers as well. For me, and also in Engelking, a net is a convergence notion (a generalisation of a sequence), so it's good to have 2 words for them. http://mathoverflow.net/questions/87026/possible-errata-in-nicolas-bourbakis-general-topology-i-chapter-1-exercise-2 Comment by Henno Brandsma Henno Brandsma 2012-01-30T14:27:31Z 2012-01-30T14:27:31Z errata = a list of errors in a published work. So it's not an &quot;errata&quot;, but possibly (at most) an error. But I don't think it is, it's correct as stated. http://mathoverflow.net/questions/86948/measure-theory-and-continuum-hypothesis Comment by Henno Brandsma Henno Brandsma 2012-01-29T10:50:15Z 2012-01-29T10:50:15Z Such sets are not Borel, but can be measurable. In case it is, it is consistent to assume Martin's axiom plus non-CH and then all such measurable X have measure 0. But it also possible to have models of non-CH where some measurable X of intermediate size has positive measure. The last remark about the Cantor set only shows that in all models we have measure 0 sets of largest possible size, which is not relevant to the question, only insofar that it shows there are always in a non-CH world intermediate size sets of measure 0. http://mathoverflow.net/questions/86812/separation-axioms/86839#86839 Comment by Henno Brandsma Henno Brandsma 2012-01-28T06:14:05Z 2012-01-28T06:14:05Z Wilansky introduced KC (all compact sets are closed) and US (every convergent sequence has a unique limit) in the paper &quot;between T1 and T2&quot;, American mathematical monthly, 74(1967), 261-264. We then have $T_2$ implies KC implies US implies $T_1$, and all implications cannot be reversed in general. http://mathoverflow.net/questions/86812/separation-axioms Comment by Henno Brandsma Henno Brandsma 2012-01-27T18:55:06Z 2012-01-27T18:55:06Z @yuan: that property is called completely Hausdorff, and is stronger than Hausdorff, not weaker. http://mathoverflow.net/questions/84703/does-every-lindelof-uniform-space-have-a-lindelof-completion Comment by Henno Brandsma Henno Brandsma 2012-01-02T18:27:45Z 2012-01-02T18:27:45Z The previous comment of mine implies that every Lindel&#246;f (regular) space $X$ has <i>a</i> uniformity that induces the topology of $X$ and is already complete. So the original poster probably means that the uniformity on $X$ is fixed and given and the question is for its (essentially unique) completion under that uniformity. http://mathoverflow.net/questions/84703/does-every-lindelof-uniform-space-have-a-lindelof-completion Comment by Henno Brandsma Henno Brandsma 2012-01-02T18:24:18Z 2012-01-02T18:24:18Z This property for (completely regular Hausdorff) spaces is called Dieudonn&#233; complete (or topologically complete) and such spaces include all paracompact Hausdorff spaces. A theorem by Tamano (1960) characterizes these spaces as: for every $p$ in $\beta(X) \setminus X$ there is a partition of unity of $X$ such that $p$ is not in the closure (in $\beta(X)$) of the support of $f$ ($X \setminus Z(f)$) for all $f$ in that partition of unity. http://mathoverflow.net/questions/80890/if-a-topological-space-x-has-aleph-1-calibre-then-it-must-be-star-countable/80938#80938 Comment by Henno Brandsma Henno Brandsma 2011-11-15T18:14:21Z 2011-11-15T18:14:21Z @Joel, indeed the productivity of calibre $\aleph_1$ spaces is well-known. http://mathoverflow.net/questions/80272/a-question-about-connectedness-in-euclidean-space/80274#80274 Comment by Henno Brandsma Henno Brandsma 2011-11-07T18:12:09Z 2011-11-07T18:12:09Z $K$ is not a subset of $U$ (as e.g. $(0,0) \in K \setminus U$). http://mathoverflow.net/questions/79435/question-about-0-dimensional-polish-spaces/79449#79449 Comment by Henno Brandsma Henno Brandsma 2011-11-01T22:19:59Z 2011-11-01T22:19:59Z Nice one; I was also looking for some base in the Cantor space as well for a counterexample. http://mathoverflow.net/questions/78378/can-we-characterize-sets-in-a-topological-space-with-the-property-that-they-have Comment by Henno Brandsma Henno Brandsma 2011-10-17T19:31:42Z 2011-10-17T19:31:42Z A closed and discrete subspace. http://mathoverflow.net/questions/77985/locally-connected-versus-locally-compact Comment by Henno Brandsma Henno Brandsma 2011-10-13T06:07:49Z 2011-10-13T06:07:49Z Not really every neighbourhood, but a local base of them. http://mathoverflow.net/questions/77905/topology-generated-by-the-collection-of-open-sets Comment by Henno Brandsma Henno Brandsma 2011-10-12T08:57:13Z 2011-10-12T08:57:13Z Cofinite is not needed, T_1 follows regardless http://mathoverflow.net/questions/76134/topological-characterisation-of-the-real-line/76139#76139 Comment by Henno Brandsma Henno Brandsma 2011-09-22T18:55:48Z 2011-09-22T18:55:48Z I like the formulation with regular better because in order to define metrisable one needs the real line, or else we need to replace it with a purely topological characterisation of metrisability (which likely will involve regularity anyway).