# Tag Info

0

By simple computations, one checks $AB=BA$, $AC=CA$, $CB=BCA$ is a presentation. So in the abelianization $A=1$, $BC=CB$ the abelianized group is $\bf Z^2$. Note that $dy, dz$ are invariant by $G$ and form a base for the de Rham co-homology

1

Getting a finite open cover $K$ consisting of metric balls seems to be a red herring. Your assumptions imply there is an open $U$ containing $K$ over which $f$ is injective. Let $\epsilon=d(K,\mathbb R ^n \setminus U)$. All you have to show is $\epsilon>0$, because clearly $\{x\in \mathbb R^n : d(x,K) < \epsilon \}\subseteq U$. Suppose ...

0

Here’s another variation on the idea of using compactifications to avoid algebraic topology: $\Bbb R\times[0,1]$ has a two-point compactification, but $\Bbb R^2$ does not. The two-point compactification of $\Bbb R\times[0,1]$ is pretty evident. Suppose that that $\Bbb R^2$ has a compactification $X=\{p,q\}\cup\Bbb R^2$, where $p\ne q$. Let $U$ and $V$ be ...

1

In a Hausdorff space, compact sets are closed. So $K_1\cap K_2$ is closed in $K_1$. Since a closed subset of a compact set is compact, you're finished.

2


2

This result is true under weaker hypothesis - connectedness (not necessarily path connectedness) of $Y$ is all we need. Fix a point $y_1\in Y$. Let $|p^{-1}(y_1)|=k$. Suppose there is some $y_2\in Y$ such that $|p^{-1}(y_2)|\neq k$ Define the set $A:=\{y\in Y\ |\ |p^{-1}(y)|=k\}$. We would like $A$ to be all of $Y$. For this we will use the connectedness ...

0

HINT: for the first question you can use the stereographic projection to see that a sphere minus a point is homeomorphic to a plane, in particular all homotopy groups are the same. For the second question: use the Van-Kampen theorem to see that $\pi_1(S)$ is trivial. The conclusion follows.

1

Fix $x_0 \in X$ and let $x_a$ be a net converging to $x_0$ in $X$. You need to show $f(x_a)\rightarrow f(x_0)$ in $\mathbb{R}$. Let $G_f$ denote the graph of $f$, which is a subset of $X \times \mathbb{R}$. Then $(x_a,f(x_a))$ is a net in $G_f$, which is compact, so we can extract a convergent subnet $(x_b,f(x_b)) \rightarrow (x,y) \in G_f$. But then $x_b ... 1 First of all note that any two distinct points in$X$can be separated by open sets. This is shown as follows - Let$y,z\in X$with$y\neq z$. Then$d(y,z)=:r>0$. Then the open balls$B(y,r/2)$and$B(z.r/2)$are disjoint open sets where the first one contains$y$and the second contains$z$. Now coming to your problem - For each$a\in A$there is a ... 1 There are three different definitions of subnet in fairly common use; their definitions are given in this question and answer. I suspect that you’re using one of the definitions given in the question; the hint below is written for the second of those but is easily adapted if you’re using one of the other definitions. HINT: Let$\nu=\langle D,\le\rangle$be ... 0 It need not be an annulus; here’s a counterexample. Let$X=\Bbb N^{\Bbb N}$, where$\Bbb N$has the discrete topology;$X$is homeomorphic to the irrationals. For distinct$x=\langle x_n:n\in\Bbb N\rangle$and$\langle y_n:n\in\Bbb N\rangle$in$X$let $$\delta(x,y)=\min\{n\in\Bbb N:x_n\ne y_n\}\;,$$ and let $$d:X\times X\to\Bbb R:\langle ... 2 Hint: By the Intermediate Value Theorem, f must be order preserving (strictly increasing). If n>1 and f^n=f, then you have f^{n-2}\circ f=f^{n-1}=\iota; thus the identity is the composition of two order preserving functions one of which is f. Now can you prove f=\iota? 1 Hint - Using the universal property of quotient spaces you can get a continuous map \phi:(X\times X)/ {\sim'}\to X/ {\sim}\times X/ {\sim} which in this case you can show to be bijective and open. 2 Hint The homeomorphism is given by$$f:(X\times X)/{{\sim}'}\longrightarrow (X/{\sim})\times (X/{\sim})$$defined by$$f([(x,y)])=([x],[y]).$$Prove it ! 0 Here is a proof: Let \gamma be a path from y_1 to y_2. For each point x \in p^{-1}(y_1), there is a unique lift \gamma_x of \gamma to a path which begins and x and ends at a point x' of p^{-1}(y_2). The map x \mapsto x' from p^{-1}(y_1) to p^{-1}(y_2) is an injection: two of these lifts can't intersect, and therefore can't end at ... 7 Any countable subset of \mathbb{R}^2 (with at least two elements :P) is indeed disconnected; but I think it takes a different argument than what you sketch to show this. Here's a proof outline. Let A\subseteq\mathbb{R}^2 be countable and fix distinct a, a'\in A. Say that a positive real r is good if r<d(a, a'), and for no b\in A do we have ... 3 Perhaps I will write something wrong? (This seems easy to me). First of all, I think you mean there can't exist an injective function f:I_m\to I_n, for if n<m there actually exists an injective I_n\to I_m (the inclusion). Suppose n<m and f:I_m\to I_n injective. Then f:I_m\to f(I_m) is bijective, where f(I_m)=\{f(1),...,f(m)\}. Clearly ... 0 Suppose \;X\; is infinite, then \;\{\{x\}:\;x\in X\}\; is an infinite open cover of \;X\; which has no finite subcover, contradiction. 1 Hint: Fix any x_0 in your domain and look at the set of x in your domain for which f(x) = f(x_0). Is this set open? Is it closed? What does that tell you? 1 Take the open unit ball B in an infinite dimensional Banach space. If B is totally bounded so is its closure (exercise!). Hence \textrm{cl}(B) is complete and totally bounded, thus compact. But this is false. 4 Start with an arbitrary metric space (with metric d) that is not totally bounded, and take the new metric \overline{d}(x,y) = \min(1, d(x,y)). 3 Let \mathscr T be a topology on a non-empty set X and let \mathscr S\subseteq \mathscr T. Then, by definition, \mathscr S is a subbasis for the topology \mathscr T if for any U\subseteq X, one has that U\in\mathscr T if and only if U can be expressed as a union of sets which are finite intersections from sets in \mathscr S. Formally: ... 1 Here's a partial result. In particular, this answers your question in the affirmative for Hausdorff spaces assuming CH, or more generally for Hausdorff spaces if you replace \aleph_1 by \mathfrak{c}. Theorem: Let X be a Hausdorff space with |X|>\mathfrak{c}. Then there is a subspace Y\subset X of cardinality \aleph_1 which is not ... 2 A connected set is a set that cannot be divided into two disjoint nonempty open (or closed) sets. Intuitively, it means a set is 'can be travelled' (not to be confused with path connected, which is a stronger property of a topological space - every two points are connected by a curve). A simply connected set (let me short it to SC for now) is path-connected ... 1 A connected set is a set that cannot be split up into two disjoint open subsets (this of course depends on the topology the set has; for the case of \mathbb{C}, this is the same as the Euclidean topology on \mathbb{R}^2). Now, a simply connected set is a path-connected set (any two point can be joined by a continuous curve) where any closed path (a ... 1 A connected set is a set which cannot be written as the union of two non-empty separated sets. This is when the set is made only of one-part, if one wants to think of it intuitively. However, simple-connectedness is a stronger condition. It requires that every closed path be able to get shrunk into a single point (continuously) and that the set be ... 0 One would definitely need some order structure, otherwise the very notion of “sides” fails to make sense. Here is an idea. Let X and Y be two topological spaces and let \succsim be a (say) partial order on X. Take any function f:X\to Y and fix x_0\in X and y_0\in Y. Let$$L_0\equiv\{x\in X\,|\,x_0\succsim x\}$$be the “lower contour set” ... 1 I propose a proof in the spirit of what you have attempted. Let us define$$g((x_1, x_2, \dots, x_{n+1})) := x_1^2 + \dots + x_{n+1}^2 -1 $$then$$M=g^{-1}((-1,1))$$Thus M is the reciprocal set of the open set \mathbb{R}, therefore an open set of \mathbb{R}^n. 1 We have the following inverse image$$ M^c=\|\cdot\|^{-1}\left([1,\infty) \right)$$the subset$[1,+\infty)$is closed in$\mathbb{R}$thus the subset$M^c$is closed in the metric space$\mathbb{R}^{n+1}$($\|\cdot\|$is a continuous function over$\mathbb{R}$), that is$M=\left(M^c \right)^c$is an open subset in$\mathbb{R}^{n+1}.$1 Consider the complement of$M$and take a sequence there to show that it is closed. Since we are working in a metric space, this approach works. 1 a) The maps$p:X\to Y$both open and locally injective you are looking for are, practically tautologically, the local homeomorphisms. An alternative terminology (especially for us ze French) is that$X$is an étalé space over$Y$. They are very important and correspond to the historical definition of sheaves, before the definition of sheaves as functors ... 3 Sure. Let$x\in M$. Put$\delta=1-\|x\|$. If we show that the ball of radius$\delta$around$x$is contained in$M,$this implies that$M$is open. The key fact is that$\|z\|=(z_1^2 +\ldots+z_{n+1}^2)^{1/2}$is a norm. In particular, it satisfies the triangle inequality. So, if$\|z-x\|<\delta$, then ... 3 For "image" this is clearly impossible, since every single point is a closed subset, its image must be a single point, and those are not open. For inverse image, constant maps have this property because the preimage of anything is either the empty set or all of$\mathbb R$, and both are closed and open at the same time. The only interesting question is, ... 1 Let$f$be an element of$C_0$, for every$c>0$, there exists$n_0$such that$\mid x\mid>n_0$implies that$\mid f(x)\mid <c/4$. Stone-Weirstrass applied to$C([-n_0,n_0])$implies that there exists a polynomial$p$such that$\|f(x)-p(x)\|<c/4$for$\mid x\mid<n_0$. You have$p=e^{-x^2}(e^{x^2}p)\$. By multiplying the analytic development of ...

0

Interior point: Let S be a set in R .x €S is said to be interior point of A if there exists e>0 such that B (x;e)subset of s

Top 50 recent answers are included