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This seems to be an extremely badly worded sentence. On p.157 of Encyclopedia of General Topology (Hart, Nagata, Vaughan), they describe Scott topology, and mention that Scott constructed "continuous lattices" that are homeomorphic to their space of self maps $[L \to L]$. Here's the relevant excerpt (I believe fair use allows me to quote a sentence from a ...

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Take $X=\mathbb{N}$ with the cofinite topology and $X_i = \{1,\dotsc,i\}$. Then each $X_i$ carries the discrete topology, and it follows easily that their colimit $\varinjlim_i X_i$ also carries the discrete topology. But the union $\cup_i X_i$, equipped with the subspace topology, actually equals $X$ and doesn't carry the discrete topology.

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You do that by induction on $k$. For $k=0$, we have $[0,1]$. Suppose that for $k$ it holds, then $A_k$ is the union of $2^k$ closed intervals, each of length $3^{-k}$. Then $A_{k+1}$ is the set generated by removing the (open) middle third of each of these intervals, the remainder is two (closed) intervals of exactly $\frac13$ of the length. So we have ...

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A pseudo metric space, i.e. a set with a metric where the distance between two distinct points can also be 0, need not be Haussdorf. For instance, consider the real numbers with an infinitesimal positive element ε, such that there's nothing between ε and 0. Then any open ball (in the pseudo metric) containing 0 will also contain ε, as d(0, ε) = 0, and hence ...

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There are lots of examples that you can come up with; see, say, Ian's in the comments. For the sake of not getting boggged down in the details, I'm going to be a little lax in my descriptions here. One important non-Hausdorff topology that comes up very naturally is the Zariski topology in algebraic geometry. It's easiest to define in terms of closed sets ...

3

We will use the Riemann criterion for complex simply connected spaces. Let $U$ be simply connected and $f$ be biholomorphic. We have only to prove that if $g:U'\to \Bbb C$ holomorphic with $g(z)\neq 0$ for every $z$, then there is a holomorphic $g_1:U'\to \Bbb C:g(z)=g_1^2(z)$ for every $z$. Let such a holomorphic function $g$.The composition $gof:U\to ... 3 The first definition makes sense in any topological vector space; the second requires a norm. In any normed linear space they are equivalent. Suppose first that$\|x\|\le M$for all$x\in E$, and let$V$be any nbhd of$0$. Then there is some$\epsilon>0$such that$V\supseteq B(0,\epsilon)$, and clearly $$E\subseteq B(0,M+1)=\frac{M+1}\epsilon ... 3 If T(x) = y, then y + \lambda^{-1} \epsilon V \subset T(x+\epsilon U). So the image of any open neighbourhood of x contains an open neighbourhood of T(x). Given y \in Y\setminus\{0\}, then v = \frac{1}{\|y\|+\epsilon} y \in V. Since V \subset \overline{T(U)}, there exists z \in U such that \|T(z) - v\| \le \epsilon \|y\|^{-1} . Let x = ... 2 After some thinking I found a counterexample. It doesn't need some very strange spaces, actually it is a rather simple construction and I'm surprised that I hadn't thought of this trick before. Let X and Y be two non-homeomorphic spaces with continuous bijections f:X\to Y and g:Y\to X. An example of such spaces is described in my question. The maps ... 2 Yet another way to see that this (or any other map) cannot be a homeomorphism is to note that, if f:S^{1}\to\left(-\pi,\pi\right] were a homeomorphism, then g:S^{1}\setminus \left\{x\right\}\to\left(-\pi,\pi\right]\setminus\left\{f\left(x\right)\right\} restricting f to the circle minus the point would also be a homeomorphism. However, ... 2 No, not even if X_1 = X_2 = X is compact Hausdorff. Take a compact Hausdorff space with cardinal strictly greater than the continuum. Then the diagonal$$ \Delta = \{(x,x): x \in X\} $$is a closed set in X \times X, therefore \Delta is a Borel set. But \Delta does not belong to the product sigma-algebra. It does not even belong to the product ... 2 For a slightly non-trivial example, consider$$f(x)=\begin{cases}\sin\Bigl(\dfrac1x\Bigr)&x\ne0,\\a&x=0.\end{cases}$$I think you will find that this function is quasi-continuous (i.e. upper and lower) if \lvert a\rvert\le1, more generally upper quasi-continuous iff a\ge -1 and lower quasi-continuous iff a\le1. 2 (Assuming you're talking about \Bbb{R}^1) Consider the open interval (a,b). Let d=|b-a|/2, and let c=(a+b)/2, the midpoint of (a,b). Take the collection of open intervals: \{(c-\frac{n}{n+1}d,c+\frac{n}{n+1}d)\}_{n=1}^\infty. This collection covers (a,b) (its union equals (a,b)), but no finite sub-collection covers (a,b). 2 I think the following may be a source of confusion: the statement "(0,1) has no finite sub cover" doesn't make any sense. You first have to choose a cover of (0,1) by open sets. Then this may or may not have a finite sub-cover. If (0,1) were compact, any such cover would (by the definition of compact in terms of open covers) have to have a finite ... 2 For a metric space \langle T, d\rangle to be complete, al Cauchy sequences must have a limit. So we add that limit by defining it to be an "abstract" object, which is defined by "any Cauchy sequence converging to it". We have two cases: The Cauchy sequence already had a limit in T. In this case there is no need to add new points, and we identify that ... 2 The question is quite weird ; the notion of simply connectedness is a topological notion and the property you are asking is a lot simpler than complex analysis. If U and U' are homeomorphic , then any topological notion on U can be carried to U' by the homeomorphism. For your question, this can take the following form. Assume that U is simply ... 2 We can take the restriction f ⊇ f': X' = f^{-1}[K] \to K. Then X' is locally compact and f' is still continuous open surjection. For every x ∈ X' take U_x its compact nbhd. Since K is compact, some finite collection \{f'[U_x]: x ∈ F\} covers K. So it is enough to take C = \bigcup\{U_x: x ∈ F\}. 2 Here is what springs to mind, though I am a bit suspicious of my proof. Suppose we have a compact set K \subset Y. For every y \in K, we may select an x \in f^{-1}(y), call this element x(y). Define J = \bigcup_{y \in K}x(y). Note that f(J) = K. I claim that J is compact. Let \mathcal C = \{U_\alpha\} over indexing \alpha be an open ... 2 What you say is true just by writing down the definition of homeomorphism and the definition of simply connected essentially. Simply connected is a topological property, and homeomorphisms let you transfer topological properties between spaces. What complex analysis does beautifully is to show that any two simply connected proper subsets of the plane are ... 2 See references in http://planetmath.org/hypergroup Besides see B.Davvaz, W.Dudek, T.Vougiouklis, A generalization of n-ary algebraic systems. Communications in Algebra, v.37(2009), pp.1248–1263, and other papers of W.Dudek. 2 For the one direction, we can use the original proof, as contained in the excerpt. Using the notation of the original, but not clinging to the formulation (which I could only very clumsily attempt to translate), Let E be a compact (limit point compact) set, and E_1 \supset E_2 \supset \dotsc \supset E_n \supset \dotsc a nested sequence of nonempty ... 2 The formula given by rewritten is correct. Although there is no addition in X, the addition in f is only happening in the D^1 component, for which addition is well-defined (as a subset of \mathbb{R}: notice that if y \in D^1 and 0 \leq t \leq 1/2, 1/2 \leq t' \leq 1, then (2t)(1-y), 2(1-t')(1-y) are again in D^1). Thus f still makes sense ... 2 Perhaps the easiest to visualise is the line with two origins which is just given by gluing two real lines together, point-for-point except for the origins of both lines. I like to think of the line with two origins as taking the real line and then 'puncturing' it at the origin but with a hole that's 'too small to see' - this is of course hand-wavey. More ... 2 There is a notion of a "group object" in a category (with products and a terminal object), which is an object (accompanied by a collection of nice maps) that simulates the behavior of a group. An equivalent way of stating this is that G is a group object in C if for every object X \in C there is a group structure on \operatorname{Hom}(X,G) with the ... 1 The set \{a,c\} is a nbhd of a because it contains an open set that contains a, namely, \{a\}. It is not a nbhd of c, because it does not contain any open set that contains c. The only open set that contains c is X itself, so X is the only nbhd of c. Another way to say it is that \{a,c\} is a nbhd of a because a is in the interior ... 1 We can see it by a few contrapositives. We can express that x belongs to the intersection of all closed sets containing S as$$ x \in \overline{S} \iff \bigl(\forall U \in \tau\bigr)\bigl( S \subset (X\setminus U) \Rightarrow x \in X\setminus U\bigr) $$by shifting the focus from the closed set C to its complement U = X\setminus C. And now we take ... 1 As the standard product topology (or weak topology) is weaker than the box topology, and hence the box topology has more closed sets, we have that$$ \mathrm{Cl}_{B}(A) \subset \mathrm{Cl}_{T}(A). $$Note also that \prod_{n\in\mathbb N}\mathrm{Cl}\, A_n is closed in both topologies, as an intersection of closed sets, and hence$$ \mathrm{Cl}_{B}(A) ... 1 Let$F_1 = [0,\frac{1}{3}] \cup [\frac{2}{3},1]F_2 = [0,\frac{1}{3^2}] \cup [\frac{2}{3^2},\frac{3}{3^2}] \cup [\frac{6}{3^2},\frac{7}{3^2}] \cup [\frac{8}{3^2},1]$and so on. The Cantor set is then$C=\bigcap^{\infty}_{k=1} F_k$. Each$x \in C$can be written in the form$ x = \frac{a_0}{3} + \frac{a_1}{3^2} + \frac{a_2}{3^3} + ...+ ...

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You are confusing homeomorphism, a bijection that is in both directions continuous, and homotopy, which is a deformation from one form to the other. In the case of knots, one needs ambient homotopies, and only the trivial knots are, by definition, ambient homotopic to the circle. Added: You can of course deform every simple knot into a circle. However, ...

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This is not at all what you're looking for, but why not look at non-Hausdorff topologies? Did you know that every finite CW complex is homotopy equivalent to a non-Haussdorff space with finitely many points? And for every subset of a space, there is a unique topology given by that set, the empty set, and the whole set. This map is certainly continuous under ...

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