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0

$\mathcal T$ and $\mathcal T'$ are not the topologies generated by base $\mathcal B$.


0

In your condition (2), it is important to point that we must have $x \in B_3$, such that $x \in B_3 \subset B_1 \cap B_2$. Let's organize our thoughts. You have fixed a topological space $(X, {\cal T})$. A collection ${\cal B}\subset {\cal T}$ is a basis if and only if satisfies conditions (1) and (2). End. Now take a set $X$, without topology, and ...


4

For the $1$-sphere (the points satisfying $x^2 + y^2 = 1$, for simplicity): Cut it into two pieces with boundaries $\{(-1, 0), (1, 0)\}$, and give each side exactly one piece of the boundary. Now your homeomorphism is simply a half rotation. The important idea here was giving each half part of the boundary. You can do something analogous with the ...


1

Here is a counterexample to the following claim: Let $X, Y$ be topological spaces with $x \in X$, and $f: X \rightarrow Y$ a function with the following property: if $M \subseteq Y$ is a connected set containing $y$, then there exists a connected set $N \subseteq X$ containing $x$, such that $M = f(N)$. Then $f$ is continuous at $x$. Let $X$ be the unit ...


0

$\newcommand{\Reals}{\mathbf{R}}$Here's a sketch using results you know about covering spaces and lifts. Express the circle $S^{1}$ as $\Reals/(2\pi\mathbf{Z})$. The idea is, if the identity map of the circle were homotopic to a constant map, then the identity map of the reals would be homotopic to a constant map through maps sending the "lattice" ...


5

Yes. Consider $$X_1=\{(x,y,z)\in S^2: y>0\vee (y=0\wedge x>0)\vee (y=0\wedge x=0\wedge z=1)\}\\X_2=\{(x,y,z)\in S^2: y<0\vee (y=0\wedge x<0)\vee (y=0\wedge x=0\wedge z=-1)\}$$ You can verify that $X_1\cap X_2=\emptyset,\ \,X_1\cup X_2=S^2$ and that $-id|_{X_1}:X_1\to X_2$ is a homeomorphism. As you can see, they are (locally and globally) ...


2

This assume that the homotopy can be defined by a smooth map. Consider $\mathbb S^1 \subset \mathbb R^2$. Then $\gamma : \mathbb S^1 \to \mathbb S^1$ can be treated as a curve in $\mathbb R^2$. Now calculate $\int_\gamma d\theta$, where $$d\theta = \frac{1}{x^2 + y^2} (-y\,dx + x\,dy).$$ When $\gamma = id$, this is $2\pi$. But for homotopic maps, the ...


1

Embedding in $\mathbb C$, the identity is $z\mapsto z^1$ and the constant map is $z\mapsto 1=z^0$. If $(z\mapsto z^n)\simeq(z\mapsto z^m)$ let $\pi_t:S^1\times I\to S^1$ be such a homotopy with $$\pi_0=(z\mapsto z^n),\pi_1=(z\mapsto z^m).$$ Lift to a homotopy $\bar\pi_t$ of paths in $\mathbb R$ starting at $0$, noting $\bar\pi_0=\bar p_n$ and $\bar\pi_1=\bar ...


2

Yes, absolutely. The approach I know about is synthetic differential geometry, which begins with thinking of manifolds not via their underlying locally Euclidean topology (this is analytic differential geometry!) but via their smooth functions, in particular those to and from the line $R$ ($R$ behaves differently enough from $\mathbb{R}$ that it's a good ...


2

As defined in Section 2.1 of Hatcher's book, a $\Delta$-complex structure on a space $X$ is a collection of maps $\sigma_\alpha : \Delta^n \to X$, satisfying a list of properties. About a page after the beginning of that definition, Hatcher gives the notation $e^n_\alpha = \sigma_\alpha(\text{interior}(\Delta^n))$, and the terminology that $\sigma_\alpha$ ...


-1

Following article is giving the proof you're looking for: Continuous functions and convergent sequences


2

No, consider $p(re^{it}) = (\sin r )e^{it}.$ Then $p$ maps $K = \overline {D(0,3\pi /4)}$ onto $\overline {D(0,1)}$ but $p(\partial K)$ doesn't even intersect $\partial (p(K)).$


0

Since $X$ is compact, there is a finite set $J\subset I$ such that $X=\bigcup_{j\in j} U_j$. Define $f:X\to\mathbb R$ by $$f(x)=\sup\{\delta>0, j\in J : B(x,\delta)\subset U_j\}.$$ Then $f$ is a continuous function defined on a compact metric space, so it attains a minimum value. Let $\varepsilon = \min_x f(x)$. That is the Lebesgue number.


1

Since $f$ is not the trivial functional it is surjective. In particular, there exists $v$ such that $f(v)=-1$. Now $C_1$ contains the $v$-translation of $\ker f$: $$ v+\ker f\subset C_1,$$ and the left hand side is dense as it is the translation of a dense set.


4

Consider a sequene $(y_n)$ of elements of $X$ with unit norm and such that $f(y_n)\leqslant -n^2$ (such a sequence exists since $f$ is not continuous). If $x$ is an element of $X$, define $x_n:=x+n^{-1}y_n$; then $\lVert x-x_n\rVert\to 0$ and $$f(x_n)=f(x) +\frac 1nf(y_n) \leqslant f(x)-n,$$ hence $x_n$ belongs to $C_1$ for $n$ large enough.


0

You aren't dealing with pre-images, so we'll have to use the definition directly. Suppose that $f$ is continuous, but that $f(\overline{A})\not\subset\overline{f(A)}$. Then exists $y \in f(\overline{A})$ such that $y \not\in \overline{f(A)}$. Then exists an open neighborhood of $y$, $V$, such that $V \cap A = \varnothing$. But $y = f(x)$ with $x \in ...


3

The usual definition of $f: X \rightarrow Y$ being continuous is that if $M \subseteq Y$ is open in $Y$, then $f^{-1}M$ is open in $X$. If you want to talk about continuity, you need to somehow be talking about open sets. You can't just talk about any old subsets. Another standard definition is that $f$ is continuous at $x$ if and only if for any open set ...


2

This seems to be a difficult problem. In the following I propose a shape that has area strictly $<{\pi\over2}$ and cannot be placed on the integer lattice without hitting a lattice point. In the figure the lattice is turned by $45^\circ$, whence $r={1\over\sqrt{2}}$. The offset $x$ is a small parameter. One computes $$a^2=r^2+x^2, \quad b^2=a^2-(r-x)^2 ...


1

Or you can follow this trace: If $x \in \overline{A} \cap B$, noticing that $B$ is open, then $B$ as the neighbourhood of $x$, $B \cap A=\emptyset$, a contradiction!


0

Note that in order that $x \in \overline{A}$ it must be the case that $U \cap A \neq \emptyset$ for every (open) neighborhood $U$ of $x$. (This is, in fact, an equivalence.) If $B$ is open and $B \cap A = \emptyset$, then $B$ is an open neighborhood of each of its elements witnessing that they are not in $\overline{A}$.


4

Since $B$ is open, $X\setminus B$ is closed. And $A\subset X\setminus B$ by assumption. Therefore $\overline{A}\subset X\setminus B$.


2

In a comment you mentioned the following correct theorem: Theorem: If $A$ and $B$ are connected and $A \cap B \neq \varnothing$ then $A \cup B$ is connected. You might have confused this with the below converse, which is false: False: If $A$ and $B$ are connected and $A \cap B = \varnothing$ then $A \cup B$ is not connected. Counterexamples have been ...


1

I think that in your reasoning there is two incorrectness: We have $\overline{X}=X\cup B\cup\{(1,\sin(1))\}$. The implication : $X$ and $B$ are connected and $X\cap B=\emptyset$ $\Rightarrow$ $X\cup B$ is not connected is false. For a counterexample one can see that $\{0\}$ and $(0,1]$ are connected and disjoint, and $\{0\}\cup(0,1]=[0,1]$ is also ...


0

I think that there is an error for $\overline{X}$. You have $\overline{X} = X \cup B$. "The curve is not oscillating on the right side"


6

The set $B$ is NOT open, so this is not a partition of $\overline X$ in open sets.


0

I have partial, but positive results. There is a fundamental book “Differential and Integral Calculus” by Grigorii Fichtenholz. This is a famous book for our students and it has many translations (but except English). I found in Appendix of the vol. I the following Theorem (I-II), with a long and complex proof. A simple curve without special points ...


3

Consider $A = \{ 2n : n \in \mathbb{N} \}$ and $B = \{ 2n-1 : n \in \mathbb{N} \}$. Note that $B = \mathbb{N} \setminus A$, and both $\sum_{a \in A} \frac{1}{a}$ and $\sum_{b \in B} \frac{1}{b}$ diverge. Therefore neither $A$ nor $B$ are convergent sets, and so by definition of the topology they are not closed. As a set is open iff its complement is closed, ...


1

It seems the following. 2,3. Let $Y$ be the set of points at which the space $X$ is locally compact. Clearly that the space $Y$ is open in $X$ and $Y$ is locally compact. By [Kech, 8.4] the space $Y$ is Baire. The set $X\setminus Y$ is $\sigma$-compact subset of the space $X$. So, if the set $X\setminus Y$ is of second category, it contains a compact subset ...


0

In fact it's and iff: If $X$ is a $T_1$ space (not even metric), then: a) $\mathcal K(X)$ is $T_2$ iff $X$ is $T_2$. b) $\mathcal K(X)$ is $T_3$ iff $X$ is $T_3$. c) $\mathcal K(X)$ is $T_{3\frac{1}{2}}$ iff $X$ is $T_{3\frac{1}{2}}$. where $T_3$ is regular hausdorff, and $T_{3\frac{1}{2}}$ is tychonoff.


3

It seems the following. I am not an expert in this question, but I think you may encounter some problems defining Haar measure on such a group. One of ways to define a (Haar?) measure $\mu$ on a non-locally compact group $G$ may be to consider a completion $\hat G$ of the group $G$. If the group $\hat G$ is locally compact and has a Haar measure ...


1

It seems the following. 1 Let $H$ be a locally compact subgroup of a Hausdorff topological group $G$. By [3.3.9] a locally compact space $H$ is open in its closure $\overline{H}$. Since $H$ is an open subgroup of the group $\overline{H}$, $H$ is a closed subgroup of the group $\overline{H}$, so $H$ is a closed subgroup of the group $G$ too. (cited from ...


3

The Tychonoff product of uncountably many compact Hausdorff spaces, each of which has at least two points, is a compact Hausdorff space which is not first countable, hence not metrizable. A group of order two with the indiscrete topology is a compact topological group which is not Hausdorff. The subspace of the real line obtained by removing all of the ...


1

$A$ is the union of all rational numbers in $(0,1)$ and isolation points of {$2,3$}. Since $\mathbb{Q}$ is dense in $\mathbb{R}$, any real number in $[0,1]$ is the limit point of $A$. Also note that no isolation point is the limit point. Denote $A^{'}$ as the limit points set of $A$. Then $A^{'}=[0,1]$ $A^o=\emptyset$ (for $\mathbb{Q}$ has no open set ...


1

You need to find those points that are outside of $A$ but still are infinitely close. By that I mean, there exists an infinite set of points in $A$ that get arbitrarily close to that point. e.g. 0 is a limit point. This is because $1/2^n$ is in $A$ for all $n$ and they get arbitrarily close. You could make the same argument regarding irrationals in ...


3

Every topological space $(X, \tau)$ has a basis: $\tau$. This is independent of said lemma.


1

By definition $\varnothing,X\in\tau$, so $\varnothing = \varnothing\cap A\in\tau_A$ and $A=X\cap A\in\tau_A$. If $U\cap A, V\cap A\in\tau_A$, then $U\cap V\in\tau$ so $$(U\cap A)\cap(V\cap A) = (U\cap V)\cap A\in\tau_A. $$ If $U_\alpha\cap A\in\tau$ then $\bigcup_\alpha U_\alpha\in\tau$ so $$\bigcup_\alpha (U_\alpha\cap A) = \left(\bigcup_\alpha ...


2

You will need an extra hypothesis to obtain a finite bound. The trouble is that if you have one loop, you can "push it off" itself to get another loop, and then you can keep pushing off more and more copies of itself to get an arbitrarily large number of loops. Of course you might say "those are the same", but that needs to be defined. A mathematican would ...


3

No. Consider $(\mathbb R,d)$ by the metric: $d(x,y)=\frac{|x-y|}{1+|x-y|}$.


4

No. The general characterization is that a metric space is compact if and only if it is complete and totally bounded. The latter means that for any $\varepsilon > 0$ the space has a finite cover by balls of radius at most $\varepsilon$ (this is sometimes called an "$\varepsilon$-net"). This rules out, for instance, the closed unit ball in an infinite ...


0

Yes, it is a covering space action, i.e. it is properly discontinuous (see the comment of @Dan). To see why, consider $p \in X$ and let $\sigma$ be the unique simplex whose interior contains $p$. The subgroup of $G$ that preserves $\sigma$ is trivial, because that subgroup stabilizes the barycenter of $\sigma$ and the action is free. So we can pick a subset ...


0

Prove that a cube and a trapezium are compact spaces. The proof that a cube is a compact space is elementary. Fist of all we prove that a segment is a compact space. Next, it is well known that there exists a continuous map $f$ from a unit segment $[0,1]$ onto a square $[0,1]^2$. Then a map $f\times f$ is continuous as a product of continuous maps, and ...


2

If $X\stackrel f\leftarrow A\stackrel g\to Y$ is a cotriad, the pushout is $$X\stackrel{\bar g}\to Z\stackrel{\bar f}\leftarrow Y$$ where $Z$ is the quotient space of $X\sqcup Y$ by the relation generated by $f(a)\sim g(a)$, and $\bar f$ is the composite $Y\to X\sqcup Y\to Z$. As often with relations, one only says what generates it, and the whole relation ...


4

There's no containment relationship between $\partial A\cap\partial B$ and $\partial(A\cap B)$ that always holds. To prove that we need two counter-examples. If $A=\mathbb Q$ and $B=\mathbb R\setminus\mathbb Q$, then $\partial A=\partial B=\mathbb R$. But $\partial (A\cap B)=\emptyset$. So it's not true in general that $\partial A\cap\partial ...


0

One point compactification of discrete topological space is a profinite space. By definition, profnite space is inverse limit of finite discrete topological space. This is equivalent to say compact Hausdorff and totally disconnected. I recommend the book "profinite groups" by Ribes and Zalesskii,page89


1

First of all, let me say that there are either open sets or closed sets in topological spaces, there are no 'half-open sets', you can have however half-open 'intervals' if the set is ordered and in the order topology these half-open intervals are open sets. Having said that, here are the answers: 1) [0,1) is an open set of $Y$ with the subspace topology. ...


1

First note that a monomorphism $f:X\to Y$ must be injective: If $f(x)=f(y)$, then the compositions of $f$ with the maps $(*\to X)$ sending the single point to $x$ and $y$, respectively, are equal, thus these maps are equal, hence $x=y$. So let $f:X\to Y$ be a monomorphism in $\mathbf{CHaus}$. Since a map from a compact space to a Hausdorff is perfect ...


2

Define $f: \Bbb R^3 \to \Bbb R$ as $$ f(x,y,z) := x^3 + y^4 - z^2 -1 \; .$$ It should be clear, that $f$ is a continous function. Note that $$ S = f^{-1}(\{0\}) \; .$$ Since $f$ is continuous and $\{ 0 \}$ is closed in $\Bbb R$, we deduce that $S$ is a closed subset of $\Bbb R^3$. It remains to show, that $S$ is unbounded. Let $x,y \geq 1$ arbitrary. Then ...


0

No, the Sorgenfrey line is not second countable. A base for the topology would have to contain a neighborhood base for each point. A neighborhood base for the point $a$ must contain some neighborhood whose least element is $a$. Therefore, any base for the topology of the Sorgenfrey line must have the cardinality of the continuum. Also see this old ...


0

Let $v\in V_{1}$. Then $\left(v,a_{2}\right)\in V_{1}\times V_{2}\subseteq f^{-1}\left(U\right)$ so that $f_{1}\left(v\right)=f\left(v,a_{2}\right)\in U$. The last statement is the same as $v\in f_{1}^{-1}\left(U\right)$ and proved is now that $V_{1}\subseteq f_{1}^{-1}\left(U\right)$.


0

It seems the following. The space $X_0$ of arbitrary functions $f : \mathbb{R} \to \mathbb{R}$ endowed with the topology of pointwise convergence is just the Tychonov product $\mathbb{R}^\frak c$. Thus in order to construct a required counterexample $X\subset X_0$ it suffices to construct a Tychonov (that is completely regular) counterexample $X$ of weight ...



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