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$(0,1)^3$ is homogeneous and $[0,1)^3$ is not, as a point on the "boundary" (like $(0,0,0)$) cannot be mapped by a homeomorphism to a point of the interior, or vice versa. It probably can be reduced to an argument involving Brouwer's fixed point theorem e.g.

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Consider the set $Y=\Bbb R^{[0,1]}$ of all functions from $[0,1]$ to $\Bbb R,$ which can be considered to be the product of $2^{\aleph_0}$ copies of $\Bbb R,$ although we use $[0,1]$ for the index of the product. A base $B$ for the (Tychonoff ) product topology on $Y$ is the set of all $\prod_{x\in [0,1]}A_x$ such that $(i). \{x:A_x\ne \Bbb R\}$ is finite,...

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Hint: Note that $[0,3)^3 \setminus \{ (0,0,0) \}$ is contractible, but $(0,1)^3 \setminus \{ x_0 \}$ is not contractible for any $x_0 \in (0,1)^3$.

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$Z\setminus\{(0,0,0)\}$ is not connected - that can only happen with 1-dimensional manifolds. But a neighbourhood of e.g. $(1,0,1)$ clearly looks 2-dimensionsl.

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It is easy to check that the metric space $c_0$ is complete. Now recall the compactness properties of subsets of complete metric spaces from [Eng] (see below). It follows that a subset $K$ of a complete metric space $(X,d)$ is compact iff $K$ is closed and totally bounded. The latter means that for each $\varepsilon>0$ the set $K$ has a finite $\... 0 Lemma: The continuous image of a compact space is compact. Little Lemma: A compact subset of a Hausdorff space is a closed set. Littler Lemma: A subspace$Z$of a Hausdorff space$X$is also a Hausdorff space. Littlest Lemma: If$f:Y\to X$is continuous and$f(Y)$has the subspace toology as a subspace of$X$then$f:Y\to f(Y)$is continuous. Theorem: If ... 2 To show that$g$is continuous, you have to show that the preimage of a closed set under$g$is closed. By definition of$g$this is equivalent to the fact that$f$maps closed sets to closed sets. Now assume that$M \subset [0,1]$is closed. Then it is compact and by continuity of$f$you get that$f(M)$is also compact. Now$X$is a Hausdorff-space which ... 1 b) and c) are quite compatible. Remember that the intersection is over all integers. If we have an orthonormal set$(x_n)$indexed by all integers and$V_j$is the closed subspace spanned by$\{x_n: n \leq j\}$then b) and c) are both true. Yes, c) means that$V_j$decreases to$\{0\}$as$j$decreases to$ -\infty$. 3 First off, it so happens that any two countable dense subsets of$\mathbb R$are "equivalent" in the sense that there is a homeomorphism of$\mathbb R$with itself that maps one to the other. Therefore, if the statement you want to prove is true, it remains true with the set$\mathbb Q$of all rational numbers replaced with any other countable dense subset ... 2 1st question. Let D = { x in B : a < x }. Let U be a nhood of s = inf D. Thus some u,v with s in (u,v) subset U. If not some x in D with x < v, then for all x in D, v <= x. As that contradicts s = inf D, exists x in D with s <= x < v. Consequently that x which is in B, is in (u,v), thus in U. 2nd question using notation from above. If ... 5 The first two bolded statements are less about using ideas from abstract topological spaces and more about using the definition of infimum, properties of real intervals and manipulating inequalities: In what follows let us denote$B_a = \{r \in B\ :\ a < r\}$. Then in every neighborhood of$s$there are points of$B$(because of the definition of the ... 4 I think the confusion comes from the notation of your open rays as$(n,\infty)$and$(-\infty,n)$. I personally find the following notation less confusing: Define${x\!\!\uparrow}=\{y\in X\mid y>x\}$and${x\!\!\downarrow}=\{y\in X\mid y<x\}$, then the set$\mathcal B=\{{x\!\!\uparrow}\mid x\in X\}\cup\{{x\!\!\downarrow}\mid x\in X\}$forms a subbase ... 1 ($\dots$) but also$A$($\dots$) Otherwise we would have$a<s-\epsilon <s\leq b'$(some$b'\in B$and$\epsilon >0)$. So$s-\epsilon$would be in$B$(because$a\in I$and$b'\in I\implies[a,b']\in I$by definition of interval ) and greater than$a$but smaller than the infimum s (last part contradicting definition of infimum). Your ... 0$(a,b)=\bigcup \{(p,q): p,q\in \Bbb Q\cap (a,b)\}.$6 If$A \subset X$let$f(x)=1$for$x \in A$and$0$for$x \notin A$. Continuity of$f$shows that$A$is open. Thus$X$has discrete topology. Hence it is homeomorphic to$\mathbb N$with discrete topology if it is infinite and to$\{1,2,...,N\}$with discrete topology if it has cardinality$N$. 2 Writing$e_m=(0,\cdots,0,1)\in \mathbb C^m$, the required map is $$G_n(\mathbb C^{m-1})\to G_n(\mathbb C^m):L\mapsto L\oplus \mathbb Ce_m=\pi^{-1}(L)$$ 1 For question 1. We may assume that$V$is a basic neighborhood (if not, then replace it with a basic neighborhood). Since$S$is a subbase of the topology (which is implied by the word "generates") then$V$is the intersection of finitely many sets, each in$S$. Say$V=V(x_1,U_1)\cap V(x_2,U_2)\cap\dots\cap V(x_n,U_n)$. Since$f\in V(x_k,U_k)$for each$...

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The only way you can test that $O$ is an open neighbourhood of the $0$-function, is that there exist finitely many $x_1, \ldots, x_n\in [0,1]$ and corresponding open intervals $U_1,\ldots, U_n$ such that $$0 \in V_{x_1, U_1} \cap \ldots \cap V_{x_n, U_n} \subseteq O$$ This is what it means that the topology on $C([0,1])$ is generated by the collection $S$ ...

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I concur with David's comment: any collection $\mathcal{S}$ of subsets of a set $X$ forms a subbase for some topology on $X$. (We generate a base for that topology by taking all finite intersections from $\mathcal{S}$, including the nullary intersection which equals $X$, etc.) But Munkres decided in his popular textbook to demand that $\bigcup \mathcal{S}=... 0 That$X$is$\sigma$-locally compact (by definition: both locally compact and$\sigma$-compact) is not enough, as shown by an example like$\beta \mathbb{N}$, the Čech-Stone compactification of the integers, which is moreover separable. I think$I^I$is an example too, BTW. If every open set is$\sigma$-compact (hence Lindelöf) it means that$X$must ... 0 For simplicity I assume that the group$G$is Hausdorff. Since$f\in (K,U)$, we have$K\subset f^{-1}(U)$. Since$\theta$is a continuous action, for each$x\in K$there exist open neighborhood$U_x$of$X$and a symmetric compact neighborhood$O_x$of$e$such that$\theta (O_x\times U_x)\subset f^{-1}(U)$. A family$\{U_x:x\in K\}$is an open cover of a ... 0 We want to know what is the cardinal of the set of subsets of$\mathbb{R}$that can be written in the form $$A=\mathcal{U}\cup C,$$ where$\mathcal{U}$is an open set and$C$is countable. Both open sets and countable sets have the cardinality of$\mathbb{R}$, so the set in question must have the same cardinality since the function sending$(U,C)\mapsto U\...

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The empty set is countable. Therefore, if $x$ is any real number then $I_x:=(x, x+1)$ is a union of an open interval and a countable set. For distinct real numbers $x, y$, $I_x, I_y$ are distinct. So the cardinality of the set that you are looking at has cardinality at least that of $\mathbb{R}$. It cannot have cardinality greater than that of $\mathbb{R}$ ...

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If I understood your notation right, you didn’t prove the key moment: why $p(U)\cap p(V)=\varnothing$. As minor remarks I note that there is no need to take preimages $m_y{-1}(U_z)$, it suffices to use compactness of a set $G\cdot y$. It seems you tend to consider $G\cdot F$ as a subset of $X$, whereas formally $G$ is a set of equivalence classes, each of ...

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You are getting they other way around, while it is true that $\exists x : P(x)$ does not necessarily imply $\forall x\in X: P(x)$ it is certainly true that $\forall x : P(x)$ does imply $\exists x\in X: P(x)$ That is to say: 2) is the case where all of the $U_i$ are nonempty, this is clearly a particular case where some $U_i$ being non-empty. (Indeed, if ...

2

Suppose that you have proven your third case. What this would mean to me is that you have shown that if you have an arbitrary collection of open sets $U_i$, some of which are non-empty, then their union is open. If some of them are non-empty (and you don't know which), and you've proven that their union is open, then you could happily assume that they're ...

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A weaker version of the claim is true: For each $x\in F$ there is $N$ such that $x\in \partial V_n$ for all $n>N$. Take some $x\in F$. Suppose $x$ does not fulfill the claim. Then there are infinitely many indices $(n_k)$ such that $x$ does not belong to the boundary of $V_n$. Since $x\in F=\cap \overline{V_n}$, $x$ is in the closure of $V_n$ for all $n$....

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Yes. We can take the countable many open intervals with rational center $\frac ab$ and radius $\frac1{b^2}$, $$U_{a,b}:=\left]\frac ab-\frac1{b^2},\frac ab-\frac1{b^2}\right[,$$ with $a\in\Bbb N_0$, $b\in \Bbb N$, $a\le b$, $\gcd(a,b)=1$. If $\frac ab\in[0,1]$ is rational, then the distance to any other rational $\frac cd$ is at least $\frac1{bd}$ and this ...

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This community wiki solution is intended to clear the question from the unanswered queue. Yes, your proof is correct.

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Writing $[0,1)$ without further comments always means that it has the subspace topology inherited from $\mathbb R$. This agrees with metric topology on $[0,1)$ induced by the metric $d(x,y) = \lvert x - y \rvert$. Note that this metric is the restriction of the standard metric on $\mathbb R$. Now you map $F$ is continuous because the two coordinate ...

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This community wiki solution is intended to clear the question from the unanswered queue. As Theo Bendit commented, you are right. You have shown that all one-point subsets are open, i.e. that $X$ is discrete. In a discrete $X$ no point is a limit point.

1

Let's show that $T \mapsto T|_D$ is a homeomorphism. Assume $T_n \to T$ in $L_1(X,Y)$ strongly. Then in particular for every $x \in D$ we have $T_nx \to Tx$ in $Y$ so $T_n \to T$ w.r.t the product topology on $Y^D$. Therefore, $T \mapsto T|_D$ is continuous. To construct the inverse of $T \mapsto T|_D$, consider $f \in Y^D$. We claim that $f$ uniquely ...

3

Let $(u_n)$ be a descending sequence of rationals converging to $a$. Let $(v_n)$ be an ascending sequence of rationals converging to $b$. $$(a,b) = \bigcup_n (u_n, v_n)\text{.}$$

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The fact that something is preserved under isomorphism (in whatever category) is the formal way of saying that it is "an inherent property of your structure".

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Define functions $f_+$ and $f_-$ from $O\to X$ as follows. For each $x\in O$ put $f_+(x)=\sup_{U} \inf_{y\in U} f(y)$ and $f_-(x)= \inf_{U} \sup_{y\in U} f(y)$, where $U$ ranges over all open neighborhoods of $x$. It is easy to check that the function $f_+$ is lower semicontinuous, the function $f_-$ is upper semicontinuous (see definitions below), and for ...

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We assume that a space $X$ is locally compact provided any its point $x$ has a local base $\mathcal B_x$ consisting of compact neighborhoods of $x$. This condition holds provided $X$ is a Hausdorff space and each point $x\in X$ has a compact subspace of $X$ containing a neighborhood of $x$ (see Munkres). Now let $x$ be any point of $X$, $f$ be any ...

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The answer given doesn't work, as it's not clear what the point $x$ or the functions $h_1, \ldots, h_n$ are. You don't seem to have constructed anything here. Note that each neighbourhood only controls the functions within it at a single point. If we consider, for example, $V_{1, (3,4)}$, the functions in this neighbourhood are precisely the continuous ...

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If I understand the question, you want to count the rearrangements of an infinite series (they would be called permutations if the series were finite). Each such rearrangement is characterized by a series of (integer) index numbers, right? So you want to know the size of the set of all series of integers, right? If nothing else, just "print out" each ...

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A better formulation would have been: the collection of closed sets of $(X,\tau)$ are closed under finite unions and also closed under arbitrary intersections. Often instead of the complete (and somewhat formal) $(X,\tau)$, the shorthand $X$ (or less commonly $\tau$) is used, which can be confusing, as we see. The fact itself is immediate by de Morgan's ...

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As Alex and José said, if there are such counterexamples, they aren't well behaved (i.e. they are not $T_0$). There is a general construction of a regular topological space which is not Hausdorff, the Partition Topology. Its description is quite simple: Given a set $X$ and a partition $\mathcal{P}$ of $X$, let $\tau$ be the topology generated by the ...

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You are correct in your statements about closed sets in $E$ with respect to the topology $t$ Once a topology is defined on $E$ we have the open sets in $E$ and their complements are closed in $E$

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Yes, it is guaranteed that the complement of any closed set in $(E,\tau)$ is in $\tau$, since, by definition, asserting that $A$ is closed means that $A^\complement\in\tau$. On the other hand, it makes no sense to assert that $\tau$ is “closed under … of closed sets”. But that is not the question. There are in fact two questions: the set of closed ...

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As correctly noted in the comments and the answer of José Carlos Santos, any regular space in which singletons are closed is clearly Hausdorff. The condition "singletons are closed" is equivalent to the $T_1$ separation axiom. So if you want an example of a regular space which is not Hausdorff, you need to look at spaces which are not $T_1$. But actually, ...

2

Typically one would take two "copies" of $\Bbb C$ and consider a quotient space of their union. But another alternative is to actually do the cutting and pasting in a higher dimension. Let me start with a figure-8 in the plane. It's got a crossing, which I'll declare "bad" (imagine it's an auto-race-course: collisions would happen there!). But if we take one ...

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There is a disparity between the definitions in the proof and the image. Let us say that red = $0$, green = $1$ and blue = $2$. According to the proof, two triangles should only be connected by an edge if the $1$-simplex joining them is labeled with $\{0,1\}$, that is, red and green. In the picture, however, we see edges between every pair of triangles whose ...

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In a space in which every singleton is a closed set (that is, in a $T_1$ space), yes, regular $\implies$ Hausdorff. But not in general. Take any set with more than one point, endowed with the trivial topology.

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As Ravsky wrote, in a normal space $X$, a set $E$ is the support of a continuous function if and only if $E$ is an open $F_\sigma$-set, and Luiz Cordeiro gave an example of a normal set $X$ and a compact set $X_1$ which is the closure of an open set, but is not $F_\sigma$, completing the answer. Moreover the claim A set $E$ is the support of a continuous ...

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Again (see previous question) the answer is no, though my example uses $m=2$. It is easily generalizable for higher dimensions, but maybe the answer is different for $m=1$. Consider the plane with polar coordinates $(r,\phi)$. Define $\forall n \ge 1$ V_n=\{(r,\phi) \in \mathbb R_{\ge 0}\times [0,2\pi): 1-\frac1n < r < 1 + \frac1n\} - \{1\}\times[0,...

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Fact: A finite subset of a linearly ordered set has a maximum. Now, if $U$ is non-empty and open it is of the cofinite form, so $\mathbb{N} \setminus U$ is finite. Define $M=\max(\mathbb{N} \setminus U)$ which exists by the first fact. If $u > M$, then $u \in U$ (or else $u \in \mathbb{N} \setminus U$ which would contradict the maximality of $M$). So ...

2

All 3 are equivalent, provided you take a correct metric on $[0,\infty) \times M$. We could take $D((t_1,x_1), (t_2,x_2)) = \max(|t_1 - t_2|, d(x_1,x_2))$ and that metric $D$ induces the product topology on $[0,\infty) \times M$. So the reformulated version of 3. would be: $f$ is continuous if for all $(t,x) \in [0 \times \infty) \times M$, and for all \$\...

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