# Tag Info

1

If $X$ is a separable metric space (that is, $X$ has a countable dense subset) and $E\subset X$ then $E$ has a countable dense subset. This is simple enough that proving it is easier than trying to look it up. Say $x_1,\dots$ is a countable dense subset of $X$. Let $$S=\{(n,j)\in\Bbb N^2\,:\,E\cap B(x_n,1/j)\ne\emptyset\}.$$For every $(n,j)\in S$ let ...

1

It is easiest to show that this is path connected. Step 1: Prove that every point in the solution set is connected to a point where $|x|=1$. Namely, if $xy=1$, then consider multiples $x/\lambda$ and $\lambda y$, but varying $\lambda$ between $1$ and $|x|$, you can continuously change $xy=1$ into a solution where $|x|=1$. Step 2: Observe that when ...

5

The set $S=\{\,(x,y)\in\mathbb C:xy=1\,\}$ is even path-connected: Given $(x_1,y_1),(x_2,y_2)\in S$, we have $x_1,x_2\in\mathbb C^\times$, which is path connected. If $\gamma\colon [0,1]\to \mathbb C^\times$ is a curve with $\gamma(0)=x_1$, $\gamma(1)=x_2$, then $\tilde\gamma\colon[0,1]\to S$, $t\mapsto (\gamma(t),1/\gamma(t))$ is a path in $S$ from ...

2

We can show the stronger (because we don't need that $G$ is Hausdorf, we don't need that $G$ is compact, and we can show that $V$ can be picked as symmetric neighbourhood of $1$) result Proposition. Let $G$ be a topological group and $A,B\subseteq G$ compact subsets. Then there exists a nonempty open set $V\subset G$ such that $1\in V$ and $V=V^{-1}$ and ...

1

You can proof that they are equivalent definitions.

2

Both definitions are equivalent. Note $$B_r(a) \subset \overline{B_r(x)} \subset B_{r+1}(x)$$

1

In fact, $x$ is a limit point of $X$ if for every open set $U$ such that $x\in U$ we have $U-\{x\}\cap X\neq \emptyset$. Thus, $\{1/n; n\in \mathbb{N}\}$ is not closed because $0$ is a limit point and $0\notin \{1/n; n\in \mathbb{N}\}$. On the other hand, $\{0\}\cup \{1/n; n\in \mathbb{N}\}$ is closed.

1

See the first paragraph after the Remark at the top of page $9$: Since $X$ is a separable metric space, by Tychonoff’s embedding theorem, $X$ is homeomorphic to a subset of a compact metric space. Thus $X$ admits an equivalent metric $\rho$ with respect to which it is totally bounded. I would appeal instead to the Uryson metrization theorem, but either ...

-1

The weak topology $\tau_{w}$ is the weakest topology for which the elements of $V^{\star}$ are continuous from $(X,\tau_{w})$ to the scalar field. For any $\phi \in V^{\star}$, the sets $\phi^{-1}B_{r}(0)$ must be open neighborhoods of $0$ for any $r > 0$. Finite intersections of such inverse images under elements of $V^{\star}$ will form a base of ...

1

The precise statement is the following: a set $A\subseteq V$ is in the weak topology iff $$\forall a\in A \exists \epsilon\in \mathbb{R}_{>0}\exists F\subseteq V^*: F\text{ is finite and }\{y\in V\mid\forall\phi\in F:|\phi(a-y)|<\epsilon\}\subseteq A.$$ It sounds like the key point you are missing is the implicit universal quantifier in the definition ...

1

This is a slight modification of an example due to K. Alster; it is Example $\mathbf{3.2}$ of J. Chaber, Remarks on open-closed mappings, Fundamenta Mathematicae ($1972$), Vol. $74$, Nr. $3$, $197$-$208$. Let $X=\Bbb R\times\{-1,0,1\}$, and for convenience set $X_i=\Bbb R\times\{i\}$ for $i\in\{-1,0,1\}$. Points of $X_0$ are isolated. For $p=\langle ... 0 Every Hausdorff topology on a real or complex finite-dimensional topological vector space is equivalent to the usual topology. So, without doing any checking: No, your topology cannot result in a TVS. 4 Scalar multiplication will not be continuous unless the topology on the scalar field$K$is also discrete (or unless$X=0$). Indeed, if$x\in X$is any nonzero vector,$a\mapsto a\cdot x$is a continuous injection$K\to X$, as the restriction of the scalar multiplication map$K\times X\to X$to the subspace$K\times\{x\}\cong K$. If$X$has the discrete ... 0 for your 2nd ques prove if F contains E then F' contains E' if E'=null set then there nothing to prove . Let E' is not a null set then xeE' implies x is a limit point of the set E in X implies E meet [B(x,r){x}] is not equal to null set for all +ve real r implies F meet[B(x,r){x}] is not equal to null set as F ... 1 The main implication we will use when proving this theorem is that: If$E\subset F $then$ E^* \subset F^*$. Proof: take$x \in E^*$, then for every neigborhood$V$of$x$, there is a point$y \in E$such that$y \in V$as well. But we have that$E \subset F$, so$y \in F$and$y \in V$. From this, we conclude that$x \in F^*$and we have the result. The ... 0 It is even true that every functionally Hausdorff connected space with at least two points has size at least continuum. In turn, every$T_3$connected space with at least two points is uncountable (a countable one would be$T_4$and hence functionally Hausdorff). On the other hand there exists a countably infinite connected Hausdorff (even Urysohn) space. 2 The spaces where every set is either open or closed are called door spaces. The spaces where the empty set and the whole space are the only clopen sets are called connected spaces. So you are asking about connected door spaces. They are in fact fully classified. There are no other such spaces then already mentioned examples – the empty space, principal ... 1 As this question is posed, it does not make much sense. We can list some classes of topological spaces for which the property $$\mbox{if X has at least two points, then X is uncountable}$$ holds. Let's make some examples. The class of connected metric spaces (as you stated) The class of spaces with one point (in this case the property is vacuously ... 0 Take$K_n:=\overline{F\cap B_n}$, then $$\lim_{n\to\infty}\lambda(K_n)=\lambda(F),$$ where$\lambda$denotes the Lebesgue measure. It should be easy to conclude from here. 1 This is definitely false, for$F = \mathbb{R}^n$you get$F \cap B_n = B_n$which is not compact. However, the closed ball$\overline{B}_n(0)$is compact and the intersection between a compact and a closed set is always compact. Therefore$\overline{B}_n(0) \cap F$is compact. Another option would be to consider$\overline{F \cap B_n(0)}$. This set is ... 0 (The OP drastically changed the question after this answer was posted.) I'm not sure if you intended the$n$in the superscript of$\mathbb{R}^n$to be the same as the$n$in$B_{n}(0)$. Nevertheless: Your question asks about the intersection of a closed and open set, so I assume that by "ball" you mean an open ball. The statement is false. Let$n = 1$... 4 Let$X$be an infinite set, and let$\mathcal U$be a ultrafilter on$X$. Then$\mathcal T = \mathcal U \cup \{ \emptyset \}$is a topology on$X$in which every subset of$X$is either open or closed, and$\emptyset$and$X$are the only clopen subsets. That it is a topology follows from the fact that$\mathcal U$is a filter, so is closed under finite ... 0 Pick any$X$and$a \in X$. Define$Y \subset X$to be open if and only i f$X= \emptyset$or$a \in Y$. It follows that a set$Y$is closed if and only if$Y=X$or$a \notin Y$. It is easy to show that this is a topology which has the required properties. It is likely that there is no separable topology with these properties. 2 Sure. $$X=\emptyset, \tau = \{\emptyset\}$$ That's cheating!!! OK, so what about if$X$is not empty? Well, here's another example: $$X=\{1\}, \tau=\{\emptyset, X\}$$ OK OK, but there is no nontrivial closed set here, that's cheating! OK then, $$X=\{1,2\}, \tau=\{\emptyset, \{1\}, X\}$$ 5 Yes, for example$X = \{0, 1\}$with the topology$\tau = \{ \emptyset, \{0\}, \{0, 1\}\}$. 4 Assume that both$A$and$\dot B:=B\setminus\{0\}$are nonempty. Then $$\dot BA=\bigcup_{b\in\dot B} b\>A$$ is a union of scaled copies of$A$, whence open. If$0\notin B$then$\dot B=B$, and we are done. If$0\in B\cap A$then$BA=\dot BA\cup\{0\}=\dot BA$as well. There remains the case that$0\in B\setminus A$. Here finer distinctions are necessary: ... 0 Results of this type, under further regularity assumptions on$E$, are established by D.Carballo "Areas of level sets of distance functions...", for almost all$\epsilon$(with estimates on the Hausdorff$n-1$-measure of the boundary of$D_{\epsilon}$). Take a look also at the earlier paper by Almgren et al, Carballo refers to. Maybe one of the references ... 1 @user87690, many thanks for the helpful hint) Maybe, one of the simplest examples is the following. On the set$\mathbb Z$of all integers with left order topology which is obviously$T_0$consider equivalence relation with partition consisting of the set of odd and the set of even integers. This relation is open and closed, since saturation of any nonempty ... 0 Another example might be$A_n=\left[\sqrt{2}-\frac{1}{n}, \sqrt{2}+\frac{1}{n}\right]\cap\mathbb{Q}$. The intersection is empty, because there is no fixed rational number arbitrarily close to$\sqrt{2}$, but each$A_n$is clearly non-empty. $$\bigcap_{n\geq1}A_n=\varnothing$$ How does this fail the hypothesis of Cantor's intersection theorem? 1 Why is there a smallest set? Consider the family of intervals$[-1,\frac1n]$for$n\in\Bbb N\setminus\{0\}$. For every interval in our family, there is a strictly smaller interval in our family of intervals. You might want to argue that$[-1,0]$, their intersection, is in the family but it is not. 1 Let$\mathcal{F_Y}=\{O:O\in X,\:O\subset Y,\:Y\in X,\: \text{O and Y are open}\}$. Then for each$O\in\mathcal{F_Y}$, there is$O=O\cap Y$for$O\subset Y$. We prove that$\mathcal{F_Y}$is a topology in$Y$. Let$O_{\alpha}$be any collection of sets in$\mathcal{F_Y}$. Then$\bigcup O_{\alpha}$is open for$O_{\alpha}$is open and arbitrary union of open ... 6 Let$\mathcal Y$be the set of all open subsets of$X$that are subsets of$Y$.. You have to check the three conditions:$\varnothing \in \mathcal Y$, which is clear and because$Y$is open in$X$, and it's a subset of itself,$Y \in \mathcal Y$. The Union of arbitrarily many elements of$\mathcal Y$is again a subset of$Y$, and each being open in$X$, ... 1 I am adding the answer I received by email from Doctor Tadashi Tokieda, he is the director of studies in mathematics at Trinity Hall, University of Cambridge, some bio here and here, I was following his Topology and Geometry open lectures at Youtube for the AIMS, so I dared to send him an email yesterday (I did not expect an answer, just tried) and received ... 2 In order to have a Suslin scheme, you must have a set$A_\sigma$for each$\sigma\in{^{<\omega}X}$; since you have only$A_\sigma$for$|\sigma|\le 2$, you don’t have a Suslin scheme and cannot define the result of the$\mathscr{A}$-operation. If you set$A_\sigma=\varnothing$for all$\sigma\in{^{<\omega}X}$of length greater than$1$, so as to have a ... 0 If$F, G$both have this property then take$n = n_F + n_G$, then$F\cup G$has no arithmetic progressions of length$n$. Proof: if it did then it would need an arithmetic progression of length$m \geq n_G$in$G$or one of length$m \geq n_F$in$F$, which is impossible. Think of gluing together finite arithmetic progressions to see this. Thus the ... 2 HINT: I would approach the union a bit differently. Let$f=f_1\cup f_2$, and suppose that$f$is not closed; then it contains arbitrarily long arithmetic progressions. Suppose that $$\{a+kd:k=0,\ldots,\ell\}$$ is an arithmetic progession in$f$, where$\ell$has yet to be determined (and remember that we can take$\ell$to be as large as we want. Let ... 4 Let$f:X \to Y$be the desired open and closed continuous surjection. Since$Y$is not$T_0$, it contains two topologically indistinguishable points. If we restrict$f$to such two-point subset of codomain, we still get a desired continuous surjection, so we may assume that$Y$is two-point indiscrete space. Hence, it is enough to find a$T_0$space and its ... 4 It suffices to consider the case of a product of two factors, so let$X,Y$be nonempty topological spaces, and suppose$X\times Y$is a$T_4$space. Let$A,B \subset X$be disjoint closed sets. Then$A\times Y$and$B \times Y$are disjoint closed sets in$X \times Y$, so there are disjoint open$U, V \subset X\times Y$with$A\times Y \subset U$and ... 1 Assume$f_n$is a Cauchy sequence in$C([0,1])$. You said you showed that$f_n(x)$is a Cauchy sequence of real numbers. The standard norm is complete on$\mathbb R$, so for any$x \in [0,1]$we know$f_n(x)$converges to some real number$\alpha_x$. Now, define a function$f$on$[0,1]$by$f(x)=\alpha_x$. We need to show that$f \in C([0,1])$and that ... 1 Since$H_i$is continuous we know that$H_i^{-1}(ClX_0)$is a closed set which contains X. Since$Y_i$is the closure of X, we know that$Y_i \subset H_i^{-1}(ClX_0)$. Therefore$H_i(Y_i) \subset ClX_0$. 0$\mathrm{Int}(M)$with the subspace topology is second countable and Hausdorff. Fix$ p \in \mathrm{Int}(M)$. There is an interior chart$(U, \phi)$with$p \in U$. From the definition of interior chart,$\phi(U)$is an open set of$\mathbb{R}^n$, which implies that$\mathrm{Int}(M)$is locally Euclidean and hence a topological$n$-manifold. If ... 0 The concept of a derivative requires a little more structure than a topological space provides. However, we can provide a derivative of a map between manifold. A manifold is a Hausdorff, second-countable topological space for which every point is contained in a neighborhood which is homeomorphic to a subset of$\mathbb R^n$(locally Euclidean). Examples of ... 0 Your guess at using$\epsilon = x_1^2 + x_2^2 - 1$is not correct, but perhaps it is going in the correct definition. Instead, you should use$\epsilon = \sqrt{x_1^2 + x_2^2} - 1$. The geometric/intuitive reason for this choice of$\epsilon$is that it is equal to the shortest distance from$x$to the set$A^c=\overline B(O,1)$: draw the segment ... 0 Let$x=(x_1,\dots,x_n)\in\mathbb{R}^n$and$||x||_3^3=\sum_{i=1}^n|x_i|^3$. Using the fact that$\forall 1\leq i\leq n$,$|x_i|\leq||x||_{\infty}$, we can chose$m=n^{-\frac{1}{3}}$. On the other hand: $$||x||_{\infty}^3=\left(\max_{1\leq i\leq n}|x_i|\right)^3=\max_{1\leq i\leq n}|x_i|^3\leq\sum_{i=1}^n|x_i|^3=||x||_3^3$$ So you can chose$M=1$. 0 We have the following: Suppose$z:=(x,y)\in R^2$. First prove that$| x |^3 + | y |^3\leq(| x | + | y |)^3$. Therefore,$2 \Vert z\Vert_\infty \geq \Vert z\Vert_1 \geq \Vert z\Vert_3$. On the other hand,$|x|^3+|y|^3\geq \Vert z \Vert_\infty^3$. Thus,$\Vert z \Vert_3 \geq \Vert z \Vert_\infty$. Summing up:$2 \Vert z\Vert_\infty \geq \Vert z\Vert_1 ...

2

It is enough to show that when one of the norms is 1, the other is bounded between two positive constants, $m$ and $M$. It is easiest to look at the case when $||(x,y)||_\infty=1$. What does that equation tell you about $x$ and $y$? You can use this to bound $||(x,y)||_3$. In terms of the shape of $||.||_3$, you really just want to sketch the curve ...

3

Here's another (probably unnecessarily elaborate) proof that they are not diffeomorphic using Eliashberg's amazing theorem about which open $2$-handlebodies admit Stein structures and the adjunction inequality for Stein surfaces. I am sorry if this is not really accessible for the asker. $M$ is the 4-manifold given by gluing an open $2$-handle along the ...

1

Let $d'$ be the metric restriced to $A$, $B_d(x,\epsilon)=\{y\in S:d(x,y)<\epsilon\}$ and $B_{d'}(x,\epsilon)=\{y\in A:d'(x,y)<\epsilon\}$ $\underline{A\cap(B\cup A^c)^\circ\subseteq B}$ Suppose $x\in A\cap(B\cup A^c)^\circ$. So $x\in A$ and $x\in(B\cup A^c)^\circ$. Hence there is an $\epsilon>0$ such that $B_d(x,\epsilon)\subseteq B\cup A^c$. ...

0

I may be interpreting your question wrong but it seems to me you are asking if $X$ is closed or open in the metric space $M=(X, d)$. And the answer to that is: yes. The entire metric space is always closed AND open within itself. This can be confusing. We know $(0, 1)$ is open in $\mathbb{R}$ but it is true that $(0, 1)$ is also closed when the metric space ...

0

Your uncountable disjoint union must have exactly $c = |\mathbb{R}|$ summands. I would visualise it as the plane $\mathbb{R} \times \mathbb{R}$ thought of as the disjoint union of $c$ vertical lines. Give each vertical line the usual topology and accept any union of vertical open sets as open. You can view this either as the product of $\mathbb{R}$ with the ...

Top 50 recent answers are included