New answers tagged

8

There even exists a largest set $X$ to which $f$ can be continuously extended. The trick is the following result (which I state here in more generality, to point out which topological assumptions one needs): Theorem. Let $X,Y$ be topological spaces, where $Y$ is $T_3$, and let $D \subseteq X$ be dense. Let $f: D \to Y$ be continuous and assume that the ...


15

It seems worth giving the cup-length argument, as it's relatively short and sweet. Suppose $\mathbb{R}P^n=U_1\cup\cdots\cup U_n$, with each $U_i\approx\mathbb{R}^n$, and let $c\in H^1(\mathbb{R}P^n;\mathbb{Z}/2)$ be the generator. For each $i$ the inclusion-induced map $H^1(\mathbb{R}P^n;\mathbb{Z}/2)\to H^1(U_i;\mathbb{Z}/2)$ is trivial, so by the long ...


1

This proof can be considered a variation of the proof using ultrafilters on $X$. I want mainly to point out that we can avoid transferring the ultrafilters through the projections if we use a slightly more general characterization of compactness using ultrafilters.$\newcommand{\FF}{\mathcal F}\newcommand{\UU}{\mathcal U}$ Definition. Let $X$ be a topological ...


34

Expanding on the comment by @user127776, the key reference is Palais, "Lusternik-Schnirelman Theory on Banach Manifolds", Topology 5 (1966), where it is proved that if $X$ can be covered by $n$ contractible closed sets, then the cup-length of $X$ is strictly less than $n$. (Here the cup-length is the largest $n$ such that for some field $F$ and ...


2

While Pietro Majer has already provided an argument and answered the questions; here is a Paper reference for a more general parameter dependent implicit function Theorem (also has Differential dependence on parameters) https://arxiv.org/abs/math/0303320 Glöckners Paper was published in the Israel Journal of math some years later. Addendum (from my comment ...


5

Assuming $P$ first countable, the standard contraction principle and elementary bounds are sufficient to conclude. You do not need higher regularity: Let $X$, $Y$ be Banach spaces, $P$ a topological space, $f:X\times P\to Y$. Assume that i. $f:X\times P\to Y$ is continuous; ii. $f(\cdot,p):X\to Y$ is differentiable, for any $p\in\ P$. iii. $f(x_0,p_0)=0$ ...


0

There are several results in topology about metrics taking values in (the positive cone of) a partially ordered Abelian group $\mathbb{G}=(G, <, +, 0)$. The first occurrence of this spaces I'm aware of is dated 1950 by Sikorski (MR0040643 - R. Sikorski, Remarks on some topological spaces of high power). He literally started a new branch of topology that ...


4

The projective space $PE$ of a topological vector space $E$ is Hausdorff but in general is not Tychonoff, not functionally Hausdorff and even not Urysohn (let us recall that a topological space is Urysohn if any distinct points have disjoint closed neighborhoods). As a suitable counterexample, consider the countale product of lines $E=\mathbb R^\omega$. The ...


10

No, the Stone-Cech compactification $\beta\mathbf{N}$ of $\mathbf{N}$ is extremally disconnected, but not the Stone-Cech boundary $\beta\mathbf{N}\smallsetminus\mathbf{N}$. To see this, it is enough to find an increasing sequence $(F_n)$ of clopen subsets with no supremum (=least upper bound) in the Boolean algebra of clopen subsets, or equivalently of the ...


3

Not an answer. I'm just expanding a comment about @PeterTaylor's observation that the known pseudovertices $X(4)$, $X(74)$, $X(1138)$ lie on the Neuberg cubic ... Bernard Gibert's "Pairs and Triads of points on the Neuberg Cubic connected with Euler Lines and Brocard Axes Isometric Parallel Chords" Proposition 1 characterizes the Neuberg cubic of $\...


3

Let $F$ be a closed subset of $[0,1]$ (say) with empty interior and $\mu(F)>0$ (where $\mu$ is Lebesgue measure), for example given by a fat Cantor set. Clearly, $1_F$ is Baire class $1$. I claim that $1_F$ cannot be almost everywhere equal to an almost everywhere continuous function. Indeed, assume that $f = 1_F$ almost everywhere, say $f(x) = 0$ if $x ...


5

For a general source of counterexamples: look at connected but not locally connected spaces. The retracts are connected but the neighbourhoods of some points are not. The Topologist's sine curve, Knaster's Bucket Handle, and the Pseudoarc are well-known examples.


8

In J. Andres, M. Väth, Calculation of Lefschetz and Nielsen Numbers in Hyperspaces for Fractals and Dynamical Systems, Proc. Amer. Math. Soc. 135 (2007), 479-487, it was shown (esssentially, the result was already implicitly shown in D.W. Curtis, Hyperspaces of noncompact metric spaces, Compositio Math 40 (1980) (2), 139-152, without explicitly noting it): $...


4

This is a report on an unsuccessful computational approach which is rather too long for a comment. I work with complex numbers to represent the points in the obvious way. It suffices to consider $\mu(z) = t(z,0,1)$ because this can be extended under the invariants to the full $t(z,z',z'')$. Since multiplication by a complex number is just rotation and ...


0

Étale spaces of sheaves provide many examples of the "not even Hausdorff" kind. In fact, starting with any sheaf of sets, you can associate an étale space such that the original sheaf is the sheaf of sections of the space.


1

Here's a very elementary example: the set $\mathbb{T}$ of triangles in the plane up to similarity can naturally be construed as a triangle itself, and hence can be identified with an element of itself. This was initially observed by Gaspar (All triangles at once), and Stewart subsequently provided a more abstract analysis of the situation (Why do all ...


4

In the Griffiths Twin Cone (or double cone over the shrinking wedge of circles), $G\subseteq \mathbb{R}^3$, all injective loops are null-homotopic yet $\pi_1(G)$ is uncountable. Hence, injective loops don't generate any of the fundamental group. However, as @WillSawin mentions in the comments, it is possible that you can pass to $G\times C$ where $C$ is ...


1

For the new version of the question (where you allow to replace the space by a homotopy equivalent one) the answer is now "yes": just replace every $X$ by $\lvert\operatorname{Sing}(X)\rvert$. This is homotopy equivalent at least for $X$ a CW complex, and since $\operatorname{Sing}(X)$ is a Kan complex, every element in $\pi_1$ is represented by an ...


10

The answer is no. Since the Hilbert cube is compact and locally contractible, such a group would be a locally contractible locally compact group. And every locally contractible locally compact group is Lie (i.e., locally homeomorphic to $\mathbf{R}^d$ for some integer $d<\infty$). For a reference Szenthe, J. On the topological characterization of ...


2

Your set is a countable dense subset of $(\mathbb{Z}/10\mathbb{Z})^\omega$; cf. Lucia's response here. Hence it is neither open, nor compact.


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