User stephen s - MathOverflow

Answer by Stephen S for Example of a weak Hausdorff space that is not Hausdorff?

2012-02-14T11:38:37Z

The one-point compactification of $\mathbb{Q}$ has the property that every compact subset is closed. So it is certainly a weak Hausdorff space. But it isn't Hausdorff, as $\mathbb{Q}$ isn't locally compact.

Addendum

Another example is the cocountable topology on an uncountable set. No two points have disjoint neighbourhoods, and the only compact subsets are the finite subsets.

Answer by Stephen S for metrizable space

2012-01-05T16:46:03Z

(This is just an expansion of my comment above.)

As Buschi Sergio points out, a topological space that is the union of countably many metrizable subspaces need not even be Hausdorff. My example of CW-complexes is intended to show that even when such a space is Hausdorff (and paracompact and submetrizable), it can still easily fail to be metrizable.

Every CW-complex is $\operatorname{F}_\sigma$-metrizable (that is, the union of countably many closed metrizable subsets). This is easy to see when there are only countably many cells, as the cell closures are compact and metrizable. But it's true in general, as I'll explain below. (I should perhaps clarify that by a "cell" of a CW-complex I mean what is sometimes called an "open cell". The cells partition the space.)

If a subset $A$ of a CW-complex is such that $A\cap e$ is compact for each cell $e$, then $A$ is closed and metrizable. (It's metrizable because it's the topological sum of the compact sets, which are metrizable.) Since each cell is $\sigma$-compact, we can express the CW-complex as a union of countably many closed metrizable subspaces of this form.

The simplest example of a CW-complex that is not metrizable consists of a single 0-cell and a countable infinity of 1-cells, forming a bouquet of circles. This is not metrizable as first-countability fails at the 0-cell.

Answer by Stephen S for Cardinality of connected manifolds

2011-06-16T18:47:55Z

A connected Hausdorff manifold with more than one point has cardinality $2^{\aleph_0}$.

Here's a proof sketch.

For each point $x$ of the manifold, let $U_x$ be an open Euclidean neighbourhood of $x$. Define a transfinite sequence of subsets $V_\alpha$ of the manifold as follows. Choose some point $y$ of the manifold, and put $V_0=U_y$. For each ordinal $\alpha$, let $V_{\alpha+1}$ be the union of $U_x$ over all $x$ such that $x$ is a limit of a sequence in $V_\alpha$. Take unions at limit ordinals.

Each $V_\alpha$ is open, and $V_{\omega_1}$ is clearly sequentially closed, and therefore closed (as manifolds are first countable), and is therefore the whole space (by connectedness). As we are assuming that the manifold is Hausdorff, sequential limits are unique, so it follows easily by transfinite induction that $V_{\omega_1}$ has cardinality $2^{\aleph_0}$.

Answer by Stephen S for A non-trivial property of all groups

2011-02-09T19:22:52Z

This is false for the infinite dihedral group $\langle a,b\mid b^2=1, ba=a^{-1}b\rangle$. No set $S$ works for $\epsilon\le1/3$, because there is always a subset with $\lceil{\epsilon|S|}\rceil$ elements that lies entirely in $\{a^n\mid n\in\mathbb{Z}\}$ , $\{a^{2n}b\mid n\in\mathbb{Z}\}$ or $\{a^{2n+1}b\mid n\in\mathbb{Z}\}$ , and none of these three sets generates the whole group.

Answer by Stephen S for Which vector spaces are duals ?

2011-02-03T09:01:22Z

If $V$ is a vector space of infinite dimension over a field $K$, then the dimension of its dual is given by $\dim(V^*)=|K|^{\dim(V)}$ . This is essentially just a restatement of what you have called the Kaplanski-Erdős theorem, because elements of $V^*$ correspond to functions $B\to K$, where $B$ is a basis of $V$, so $\dim(V^*)=|V^*|=|K^B|=|K|^{\dim(V)}$ .

To answer your "precise question": Let $V$ be a vector space of dimension $\aleph_0$ over $\mathbb{Q}$. Then $|V|=\dim(V)$, yet $V$ is not isomorphic to the dual of any vector space, as its dimension is not of the form $|\mathbb{Q}|^\lambda$ for any infinite cardinal $\lambda$. Moreover, the same is true if $\aleph_0$ is replaced by any other strong limit cardinal, such as $\beth_\omega$.

Note that $\mathbb{R}^{(\mathbb{R})}$ is isomorphic to the dual of $\mathbb{R}^{(\mathbb{N})}$.

Answer by Stephen S for profinite spaces coming from profinite groups

2011-01-20T17:32:49Z

(Note: This was intended to be a comment to unknown (google)'s answer - but as I'm new here I can't post comments.)

As Pete L. Clark points out, unknown (google)'s answer is false as stated. However, this is only because of the omission of the word "infinite". A correct statement is:

An infinite profinite group $G$ is homeomorphic to $\{0,1\}^{w(G)}$, where $\{0,1\}$ is the $2$-point discrete space, and $w(G)$ is the weight of $G$.

This is Theorem 9.15 (pages 95-98) of "Abstract Harmonic Analysis I" by Edwin Hewitt and Kenneth A. Ross. (Hewitt and Ross actually state the result using the minimum cardinality of a local base at $1_G$, rather than the weight of $G$, but these are equal for infinite profinite groups.)

Notice that the case of countable weight is an immediate consequence of the usual characterisation of the Cantor set.

Comment by Stephen S

2012-02-17T22:11:45Z

This is true if the group is finitely generated, but it's false in general (e.g., free groups of infinite rank).

Comment by Stephen S

2012-02-15T09:05:31Z

@Bob Solovay: Yemon Choi's definition of the one-point compactification is correct. @Yemon Choi: But your second comment is wrong, since $\mathbb{Q}$ has infinite compact subsets (e.g., any convergent sequence together with its limit forms a compact set). But every compact subset of $\mathbb{Q}$ has empty interior, so it's true that every neighbourhood of the point at infinity meets every non-empty open set.

Comment by Stephen S

2012-01-05T13:11:22Z

Every CW-complex is the union of a sequence of closed metrizable subspaces, but there are many examples of CW-complexes that aren't metrizable (a wedge of countably infinitely many circles being perhaps the simplest).

Comment by Stephen S

2011-10-11T08:16:29Z

Regarding the parenthetical "finitely generated": all polycyclic groups are finitely generated.

Comment by Stephen S

2011-04-17T08:22:01Z

@Giuseppe: If the group isn't locally Euclidean, then the subgroup needn't be discrete. For example, let $G$ be a direct product of infinitely many circle groups. The circle group has a subgroup of order $2$, so $G$ has a subgroup that is a direct product of infinitely many $2$-element discrete groups. This subgroup is closed and totally disconnected but not discrete.

Comment by Stephen S

2011-03-08T21:17:33Z

@Anton Petrunin: Finite dimension can't be right, since every compact metrizable space is a quotient of the Cantor set, and that includes things like $[0,1]^{\aleph_0}$ .

Comment by Stephen S

2011-02-21T19:19:41Z

$\mathbb{R}$ and $\mathbb{R}^n$ are vector spaces of the same dimension (namely $2^{\aleph_0}$) over $\mathbb{Q}$. So they are isomorphic as $\mathbb{Q}$-vector spaces, and therefore as groups.

Comment by Stephen S

2011-02-05T09:16:36Z

Reminds me of boustrophedon.

Comment by Stephen S

2011-02-04T09:19:29Z

@Andres Caicedo - I'm not claiming that $\beth_{\omega_1}^{\aleph_0}>\beth_{\omega_1}$. I only need that $\beth_{\omega_1}\cdot{\aleph_0}=\beth_{\omega_1}$ (to get $|V|=\dim(V)$) and that there is no cardinal $\lambda$ such that $\aleph_0^\lambda=\beth_{\omega_1}$ - and these statements are true, and don't involve $\beth_{\omega_1}^{\aleph_0}$.