# Tag Info

## New answers tagged general-topology

2

It is almost correct except that you should only have $U\subset W$ as $W$ can be big. Then it suffices to find a $V$ in $U$. In this case you might want to find $\epsilon$ so that $$\overline{B(\varphi(p),\epsilon)} \subset \varphi (U)$$ instead of just $B(\varphi(p),\epsilon) \subset \varphi (U)$.

0

$\ell^1$ is a metric space and its closure equals to the set of all elements that have distance $0$ from $A$. On one side, any element $g$ of $$B:=\{f\in\ell^1:\,\,\forall\,\,n\,\,|f(n)|\leq \frac{1}{2^n}\}$$ has distance $0$ from $A$, because it either in $A$ or a limit of elements in $A$. On the other side, $\ell^1\setminus B$ is open (being equal to ...

0

Let $\varepsilon > 0$, so that $[0,\varepsilon) = V \in \mathcal{V}_{f(1)}$. It suffices to show the condition of $V$ of this form. Note that $f(1) = 0$, so there is no problem to the left of $x$ as there $f$ only assumes $f$. So take $W = (0, 1 + \varepsilon) \in \mathcal{V}_1$. Then if $x \le 0$ we know that $f(x) = 0 \in V$ and also if $x > 1$, ...

1

$A$ is not necessarily closed, but that does not matter: any sequence has a cluster point in the compact metric space $E$ (which is all we need). Using your $r$ we indeed get infinitely many $p_i$ that are in $B(p, \frac{r}{2})$ so for some large enough index $n$ we indeed have that $B(p, \frac{1}{n}) \subseteq B(p, \frac{r}{2})$. But we need (for a ...

0

As you use the tag "metric spaces" we can assume we are in a metric space. But then the statement is quite trivial, as all sets of the form $\{x\}, x \in X$ are closed, and we can write any set $A$ as $\cup \{\{x\}: x \in A \}$. So nothing about openness and almost nothing of the metric is then used. It is true that every open set in a metric is the ...

0

So we have $V_1 = V_{f_1, k_1, \varepsilon_1}, V_2 = V_{f_2, k_2, \varepsilon_2}$, and we need to show that for any $f_3 \in V_1 \cap V_2$ (we cannot choose $f_3$, it can be any point in the intersection) some $V_3 \in \mathcal{B}$ exists with $f_3 \in V_3 \subseteq V_1 \cap V_2$. To get $f_3 \in V_3$ it's easiest to choose $V_3 = V_{f_3, k_3, ... 1 For$\textbf{part 2}$: The linear algebra is a bit messy, but you should get that$\ker \partial_1 = \langle a,b,c,d-i,d-h,f-g,e-f,d-e \rangle$and$\text{Im } \partial_2 = \langle 2a-d+f,2b-f+h,2c+d-h, -a+c+2d-e-i, -d+2e-f,a-b-e+2f-g \rangle$But, what's important is that$\partial_2(S_1 + S_2 + \ldots + S_6) = (a+e-d) + (a+f-e) + (b+g-f) + (b+h-g) + ...

1

According to my Differential Geometry professor, it means that the closure of $V_{\alpha}$ is contained in $U_{\alpha}$. According to Silvia Ghinassi and other sources, it generally means that the closure of $V_{\alpha}$ is a compact subset of $U_{\alpha}$, in which case the notation $V_{\alpha}\Subset U_{\alpha}$ is read "$V_{\alpha}$ is compactly ...

1

Warning: This answer is not complete, as it contains a claim that has yet to be proved. Wlog we may assume that $M$ is not compact. Let $K \subset M$ be a given compact subset. Then, there exists a compact, connected set $L \subset M$, so that $K \subset int(L)$. For example, one could pick a complete Riemannian metric on $M$ and choose $L$ to be the ...

1

Because $f$ is supposed to be $\mathcal{C}^1$ and $D_f(0)=I$, one can find a neighborhood of the origin $V$ such that for $x \in V$: $\Vert D_f(x) \Vert \le B=\frac{3}{2}$. Using the Mean Value Theorem in several variables, you get $$\Vert f(x)-f(y) \Vert \le B \Vert x-y \Vert$$ for $x,y \in V$. The other inequality is obtains using the Inverse function ...

0

Yes, this set is closed. In fact, we do not need all of your assumptions, for example the assumption $A\cap B=\emptyset$ is superflous. Let $\left(X_{n}-Y_{n}\right)_{n\in\mathbb{N}}$ be a sequence with $X_{n}\in A$, $Y_{n}\in B$ and $X_{n}-Y_{n}\xrightarrow[n\to\infty]{}Z$, where convergence is in $L^{p}$. We have to show $Z\in A-B$. To this end, observe ...

0

Well, the $A_i$ have a topology they inherit from $X_i$ as a subspace. Open sets in $A_i$ are exactly those of the form $O \cap A_i$ where $O$ is open in $X_i$. The product topology on $A = \prod_{i \in I} A_i$ is defined as usual: the basic open sets are those of the form $\prod_{i \in I} U_i$, where there exists a finite subset $F$ of $I$ such that for ...

1

By applying a rotation to everything, we may assume $z_0=1$. Now consider the map $g:\mathbb{R}\to[0,\infty)$ given by $g(t)=|1-e^{it}|^2$. We can compute $g(t)=(1-\cos t)^2+\sin^2 t=2-2\cos t$. It now follows easily from the fact that $\cos t$ is monotone on $[0,\pi]$ and $[-\pi,0]$ that if $0<\delta<4$, then the set ...

1

Your proof looks good to me. An outline for an alternative proof: define $g\ :\ \Bbb R \times \Bbb R^k \to \Bbb R^n\ :\ (t, x) \mapsto f_t(x)$. $g$ is then continuous. $T := [-1,1]\times S^{k-1}$ is compact. Since for all $t, f_t$ is injective and $f_t(0) = 0$, we know that $0 \notin f_t(S^{k-1})$, so $0 \notin g(T)$. Since $g(T)$ is compact, it is closed, ...

1

You have done almost all the work. Using your notation etc. Since $C\cup D=M=\cap \{B\,|\, B\in \mathscr{B}\}$ it follows that $\cap \{B - C\cup D\,|\, B\in \mathscr{B}\}=M - (C\cup D) = \emptyset$ and hence $\cap \{B - U\cup V\,|\, B \in \mathscr{B}\}=\emptyset$, but since $\mathscr{B}$ is simply ordered and each $B-(U\cup V)$ is closed it follows by ...

1

The Heine-Borel and the Bolzano-Weierstrass theorems are two fundamental results in real analysis. These theorems are equivalent in the sense that their proofs can be derived from each other. In fact, there are other axioms and results such as completeness axiom, the nested interval property, the Dedekind cut axiom of continuity and Cauchy’ s general ...

0

Hint: This space is contractible, can you see the contraction? What does this say for the fundamental group?

2

Let's start with a definition: Let $(X,\mathcal{T})$ be a topological space. Let $I$ be an index set, and let $Y_i (i \in I)$ be topological spaces and let $f_i: X \rightarrow Y_i$ be a family of functions. Then $\mathcal{T}$ is called the initial topology with respect to the maps $f_i$ iff $\mathcal{T}$ makes all $f_i$ continuous. If $\mathcal{T}'$ is ...

3

observe that $\mathbb{R^2}$ is the universal cover of torus, and since $f^*$ is the zero map, so by map lifting lemma, you can lift $f$ in $\mathbb{R^2}$, and since $\mathbb{R^2}$ is contractible, so image is contractibe , i.e image is homotopic to zero. Now compose the homotopy with covering map will give a null homotopic map in $\mathbb{T^2}$.

5

It is not true in general that a map that induces 0 on the level of fundamental groups is null-homotopic. (Take the identity map of $S^2$!) You need further argument, such as: Because $f_*$ is zero, you can factor $f$ through the universal cover $\Bbb R^2 \to T^2$. Because $\Bbb R^2$ is contractible, the map is null-homotopic. E: The same argument works ...

0

Heine-Borel and Bolzano-Weierstrass theorems are equivalent in the sense that their proofs can be derived from each other. In fact, there are other axioms and results such as completeness axiom, the nested interval property, the Dedekind cut axiom of continuity and Cauchy’s general principle of convergence which are equivalent to these theorems, stated as: ...

1

The Euclidean topology $\mathcal{T}_e$ (the planar topology restricted to $S^1$) is a specific topology that makes both projections continuous. So if we by $\mathcal{T}$ denote the smallest topology that makes both of them continuous, we by definition have $\mathcal{T} \subseteq \mathcal{T}_e$. If $(c,d)$ and $(e,f)$ are open intervals in the reals, then ...

1

The case $d = 0$ is uninteresting, since $\mathbb{R}^0 = \{0\}$ has diameter $0$. So we assume $d \geqslant 1$. Furthermore, if $L$ is a straight line in $\mathbb{R}^d$, then the orthogonal projection $\pi$ to $L$ is Lipschitz-continuous with Lipschitz constant $1$, hence $\pi(B)$ is a connected subset of $L \cong \mathbb{R}$ (where $\cong$ means an ...

0

This is also proved in Topology and Groupoids (as it was in the 1968 edition, "Elements of Modern Topology"); this has some pictures of the crucial mapping cylinder construction $M(f) \cup X$ which, if $i: A \to X$ is a cofibration, is a useful model of the adjunction space $B \cup _f X$ for $f: A \to B$. Here is a coloured picture of the homotopy as Fig ...

1

Yes, the elements of the base are considered open. In the case of the base of a specific topology this means $\mathcal B\subseteq \mathcal T$, indeed. Because the base sets are open we have not only that all elements of $\mathcal T$ are unions of base sets, but also that any union of base sets is in $\mathcal T$. No, in general $\mathcal B$ is not a ...

1

You need to use a cover of the real numbers that consists of bounded sets only, like the one suggested by Mirko. Compactness implies that a finite number of these bounded sets cover $A.$

1

You can just take an example with empty interior and add on a disjoint set which is the closure of an open set. For instance, you can take $\{0\}\cup[1,2]\subset\mathbb{R}$.

4

No, the complement of an unbounded set need not be bounded. For instance, if $A = [0,\infty)$ then $\mathbb R \setminus A = (-\infty,0)$ is also unbounded.

1

Your interpretation of the set at the end of your post is correct. The key to solving this problem is: $\star$any open interval centered at a point will contain both rational and irrational numbers. Isolated points: By $\star$ the points in $[0,3]$ are not isolated. The set of isolated points is $\mathbb{N} - \{1,2,3\}$. To show this, just take $\delta = ... 1 You're right that there's no finite subcover. But that's not a problem, because it's not a cover at all! The union of all the sets$(1/n,1-1/n)$is only$(0,1)$, not$[0,1]$. The points$0$and$1$are not in any of your open sets. 3 It is the largest topology on$\mathbb R $: every set is open (and thus every set is closed). In this topology, any function$f:\mathbb R\to T $, for any topological space$T $, is continuous. The other extreme is the topology$\{\mathbb R,\emptyset\} $, in which no nontrivial set is open, and so the only continuous functions are the constants. 1 To visualise this when$n = 2$, observe that the action of$G = \Bbb{Z}$identifies each point$\mathbf{v} \in \Bbb{R}^2 - \{0\}$with all the points$2^i\mathbf{v}$where$i \in \Bbb{Z}$. Each point in the quotient space$(\Bbb{R}^2 - \{0\})/G$is represented by exactly one point$\mathbf{v}$such that$1 \le \| \mathbf{v} \| < 2$. Thus the quotient ... 0 Hint:Remark that$S^1=R-0/<h(x)=2x>$. Consider$f:R^n-\{0\}\rightarrow S^{n-1}\times S^1$defined by$f(x)=(x/\|x\|,[\|x\|])$where$[\|x\|]$is the class of$\|x\|$in$R-0/<h>$show that$f$induces a homeomorphism from$R^n-0/<\Theta>\rightarrow S^{n-1}\times S^1$. Let$(x)\in R^n-0$, remark that$f(2x)=f(x)$thus$f$factors by a map ... 0 As I said in my comment, it suffices to consider the case where$X$is the reals.For each$p$in the set you denote as$Z$, let$f(p)\in Q^+$where $$\forall r\in ]x,y[\cap ]-f(p)+p,f(p)+p[ (r\ne p\to \psi (r)<\psi (p)).$$By contradiction,suppose$Z$is uncountable. Let$q_0\in Q^+$such that$f^{-1}\{q_0\}=\{p\in Z :f(p)=q_0\}$is uncountable. Such a ... 1 Let's use standard notation$\partial$for boundary. Since$B$is close set,$\bar{B}=B$and$B^o=(-\infty, x)$. So$\partial B=\bar{B}-B^o=\{x\}$. Since$P(X\leqslant x)$is continuous at$x, $$\lim_{\Delta x\to0}P(X\leqslant x+\Delta x)=P(X\leqslant x)$$ And thus \begin{align} P(X \in \partial B)&=\lim_{\Delta x\to0}P(x<X\leqslant x+\Delta x) \\ ... 2 The notion of a metric is supposed to capture the main features of the everyday idea of distance between two points. The most important parts of this idea are that a point is at zero distance from itself; distinct points are a positive distance apart; the distance from a pointx$to a point$y$is the same as the distance from$y$to$x$; and the ... 3 First I'll consider your (bolded) question about taking only discs with centers in$S^1$. Indeed we can do this - consider your open cover$\{ U_i\}$. Each$x\in S^1$is contained in some$U_x\in\{ U_i\}$, and$U_x$is open, so it contains some open ball in$\mathbb{R}^2$centered at$x$, say$B_x$. Indeed, it should be clear that the collection of open ... 2 Let$\varepsilon >0$and let$x\in \mathbb{R}^N$, then for any$y\in B_{\varepsilon}(x)$you have a homeomorphism $$\tau_y^{x,\varepsilon} \colon B_{\varepsilon} (x) \to B_{\varepsilon} (x)$$ which maps$x$to$y$and extends to the identity on the boundary, hence is can be extended to$\mathbb{R}^N$and we will call this map$T_y^{x,\varepsilon}$Let ... 1 First off, if$E$is dense in$V$, then$E=V$since$E$is closed. (Not clear from your post if you noticed that, just pointing out.) Now (in general) for any$x$do: (a) if$x>\sup E$then let$f(x)=f(\sup E)$(just as you already did, (b) if$x<\inf E$then let$f(x)=f(\inf E)$(note also that instead of$\inf$or$\sup$we could write$\min$or ... 0 From what Daniel Fischer wrote ,distance is symmetric . So, when you choose any arbitrary point$z\in X$and choose an arbitrary nbd$\mathcal N$of$z$. then$\existsn$such that$B(z,{1\over n})\subset \mathcal N$. Now take the$1\over n$balls of$X$that cover$X$.$z$must be in one of the balls , say$B({x_i}_n,{1\over n})$Then notice ... 3 Just show that$\{x_n:n\in\mathbb{N}\}$is discrete, i.e. every singleton is an open set. For each$n\in\mathbb{N}$, show that$B(x_n,\epsilon)$only contains$x_n$. 0 I think I have a counterexample. Let$r\to K(r)$be a differentiable real function that takes the value 0 outside the interval (0, 3), the value 1 inside the interval [1,2] and values between 0 and 1 everywhere, and such that the absolute value of its derivative never exceeds 2. Such functions exist. Now consider $$f(x,\theta)=x.\left(\theta + ... 6 I see this as an additive/multiplicative notation issue, exactly as you say. It's like when one talks about exact sequences of abelian groups or modules, one often writes$$ 0 \to A \to B \to C \to 0 $$whereas one may more likely write$$ 1 \to X \to Y \to Z \to 1 $$for an exact sequence in the category of all groups. 4 From the comment: (1) Find a continuous map f : \mathbb S^1 \times \mathbb S^1 \to \mathbb S^1 so that f_* is surjective. (2) Let \gamma : \mathbb S^1 \to \mathbb{RP}^2 represent the generator of \pi_1(\mathbb{RP}^2). Then \gamma_* is surjective. (3) The composition \gamma \circ f is one of the example as (\gamma\circ f)_* = \gamma_* ... 0 A_1=f(X).| A_2=f(A_1) . Now$$f:X\rightarrow X$$and we have$$A_1\subset X\\or,\ f(A_1)\subset f(X)=A_1\\i.e.\ A_2\subset A_1\\or,\ \ f(A_2)\subset f(A_1)\\i.e.A_3\subset A_2\\.\\.\\.\\.\\.\\so\ \ \ \ on.$$Assume that this holds upto the integer n. i.e. A_n\subset A_{n+1}. Then we see$$f(A_n)\subset f(A_{n+1}\\i.e. A_{n+1}\subset ... 0 There are only countably many points$z\in{}]x,y[$such that this doesn't hold. Indeed, for any particular$Z$, there is at most one$z\in Z$such that the condition fails for$z$on$Z$, namely the point of$Z\cap{}]x,y[$where$f$takes its unique minimum value (if such a point exists). Now let$\{Z_n\}$be a countable basis for the topology on$]x,y[$. ... 2 If you don't want$f$to be a quotient mapping, then you can easily construct a meta-example, in the spirit of @bof's example. Let$Y$be your favourite Hausdorff, non-metrisable space, and let$X$be$Y$with discrete topology. Then the "identity" mapping$X\to Y\$ works. Of course, it will never be a quotient mapping. As an aside, what you are ...

1

We may define i: I → I to be the identity map, i¯: I → I to be i¯(s) = 1-s and e: I → I to be e(s) = o for all s∈I. Then we may define H: IxI → I by H(s,t) = (1-t)(i∗i¯)(s)+te(s) for all (s,t)∈IxI. Since I is a convex subset of real numbers, H is well defined. Then you may check f。H is a path homotopy between f∗f¯ and ex0.

1

Enumerating all of the elements using the diagonalization technique allows you to pair every element of your collection to a unique natural number. If you decided to first try and pair off all the elements of your first countable set with the natural numbers, you would never reach a point where you could start indexing the second set.

1

Start with the (excellent but outdated) handbook of set theoretic topology and then move onto (recently revised) open problems in set theoretic topology. This has evolved into a vast subject so it is hard to recommend specific papers too strongly. My advice is: Learn the basic techniques (elementary submodels, forcing, combinatorial principles), fall in love ...

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