# Tag Info

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HINT: Finite sets are always compact. But they can be dense, if you don't require $T_1$.

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It would be like $\Phi(x_1,\dots x_n)=(x_1,\dots,x_n,1/f(x_1,\dots,x_n))$, namely $x_{n+1}*f(x_1,\dots,x_n)=1$

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It means the first. The support of a function $f$ is the collection of points in the domain that do not map to $0$. To say the support lies in $V$, we mean the support is a subset of $V$. Now if the function you choose is the $0$ function, there is never a non-empty $A$ so that $f(A)\neq 0$ (I assume, you implicitly mean, $\forall a \in A (f(a)\neq 0)$. ...

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Since $X$ is second countable, all you need for $X^*$ to be second countable is a countable neighborhood base for the point at infinity, right? Since $X$ is locally compact and second countable, you can cover $X$ with countably many open sets whose closures are compact. Then every compact subset of $X$ is contained in a finite union of those open sets, and ...

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Try to establish the following facts: $X$ is $\sigma$-compact. $X$ is hemicompact. Use the hemicompactness of $X$ to construct a countable neighborhood base at $\infty$ in $X^*$. Conclude that $X^*$ is second countable. See Theorem 3.44 in Aliprantis–Border (2006, pp. 92–93) for more details.

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If $d$ is a metric of space $X$, the the conclusion is right. Since $d$ is continuous and the set is equal to $d^{-1}([r,s])$. Note that for a continuous function, any inverse images of closed sets are closed.

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Fixed $x_0 \in X$, then $f(y)=d(x_0,y)$ is a continuous function( you can actually show it is Lipschitz continuous). Then your set is just preimage of a closed set of $\mathbb{R}$ under a continuous function, which is closed. Or you can show its complement is open, which is a union of two open sets.

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In general, you can define a map from $X$ to $A/B$ by defining a map from $X$ to $A$, and then taking the equivalence class of the resulting element of $A$. But not every map from $X$ to $A/B$ can be obtained this way -- indeed, many of the interesting ones turn out not to be. (To be more exact: if you see a continuous map from $X$ to $A/B$, it may not arise ...

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Any continuous function valued in $D^n$ extends to your quotient space. This is actually simply because the composition of continuous functions is continuous: you have a quotient map $D^n\to D^n/\sim$, and so you get a continuous map $S^n \to D^n/\sim$ as the composition $S^n\to D^n\to D^n/\sim$.

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Let $\kappa$ be an infinite cardinal. Let $X$ be a set of power $\kappa$, and let $\tau$ be a topology on $X$. There are $2^\kappa$ equivalence relations on $X$, so $\langle X,\tau\rangle$ has at most $2^\kappa$ distinct quotients. For each $p\in\beta\kappa$ let $Y_p=\{p\}\cup\kappa$ with the topology that it inherits from $\beta\kappa$. For ...

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No, in general this is wrong! Consider the Borel algebra generated by the cofinite topology: $$\Sigma:=\{E\subseteq[0,1]:\#E\leq\aleph_0\lor\#E^c\leq\aleph_0\}=\sigma(\mathcal{T}_{cofinite})$$ and the finite Borel measure: $$\lambda(\#E\leq\aleph_0):=0$$ $$\lambda(\#E^c\leq\aleph_0):=1$$ Then the atoms are all the uncountables but no singleton is a point ...

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All three points you suspect choice are using choice. Although in the third one you can perhaps remove it. There might be many finite subcovers at each step, and you have to choose one each time. How do you choose it? There might be many candidates at each time. How do you choose them? Ideally, you have a finite enumeration of $Q_n$, and you choose from ...

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The example you give actually is a $\sigma$-finite Borel measure. Equip $[0,1]$ with the cofinite topology (in which a set is open iff it is either empty or its complement is finite). Then your $\Sigma$ is the Borel $\sigma$-algebra of the cofinite topology (it is a nice exercise to verify this). However, there is the following result: Proposition. Let ...

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Say U and V are in T , to show that the intersection belongs in T you look at it's complement , since the complement of U and V are both FINITE points and lines then the complement of $U \bigcap V$ is the intersection of finite points and lines then it would be a bunch of points since two lines intersect in a point and the lines that are common of both ...

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The inverse-image of a set $S$ under a function $f$ is the set $\{x : f(x)\in S\}$. This exists regardless of whether the inverse function $f^{-1}$ exists.

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We say that $f:X\rightarrow Y$ is a continuous map if given any open set $U\subset Y$, then $f^{-1}[U]$ is an open set in $X$. This isn't the inverse map we are using, but a related notion called the "pre-image": Let $f:X\rightarrow Y$ be a function, and for $B\subset Y$, define the pre-image of $B$ to be $$f^{-1}[B]=\{x\in X \mid f(x) \in B\}$$ This ...

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Here is a nice example I recently discussed in lecture. Recall that a function $f:\mathbb R\to \mathbb R$ is Baire one iff it is the pointwise limit of a sequence of continuous functions. These functions do not need to be continuous, but Baire proved that they always have uncountably many points of continuity; in fact, the set of points of continuity of $... 0 It can be used to (easily) show that the rational numbers$\mathbb{Q}$are not a$\mathrm{G}_\delta$-subset of$\mathbb{R}$. This in turn can be used to show that there does not exist a function$f: \mathbb{R} \to \mathbb{R}$which has exactly the set of rational numbers as its points of continuity! 1 In the course I learned this in, one of the examples (with a sketch proof) was that there exists a continuous function that is nowhere differentiable, as in Henno's comment. In a different course, we proved this by constructing an explicit example, which was certainly much more complicated (even if more elementary) than fleshing out the sketch proof. We ... 1 Let the polyhedra be$P_1,P_2 \subset \mathbb{R}^3$. By translating them we may assume that the interior of each contains the origin. Let$S$be a sphere centered on the origin whose interior contains each of$P_1,P_2$. For$i=1,2$, project the vertices and edges of$P_i$from the origin out to$S$to obtain a cell decomposition of$S$denoted$\Sigma_i$. ... 1 If$(K_i)_{i \in \mathbb{N}}$is a sequence of closed sets in$\mathbb{R}^3$, then the union of these sets$\bigcup_{i=1}^\infty K_i = K_1 \cup K_2 \cup ... $is also closed. The above statement is, as you suggest, false. Let$(K_i)_{i \in \mathbb{N}}$be a sequence of closed sets in$\mathbb{R^n}$. Then$\cup_{i=1}^\infty K_i$is closed iff ... 0 You are on the wrong trace. Generally, the union of infinity many open sets is still open, however, the union of infinity many closed sets may not closed. 1 For all$b\in B$, since$f(A)$is dense, then every neighborhood of$b$intersects$A$. Since$A$is compact,$f$is continuous,$f(A)$is compact, hence closed. Thus$b\in f(A)$otherwise we have an open set$G$containing$b$s.t.$G\cap f(A)=\emptyset$, which contradicts your assumption. So your statement holds. But please be careful that continuous ... 0 This is not true: $$(-5,5) = \bigcup_{n=1}^\infty[-5+\tfrac{1}{n}, 5-\tfrac{1}{n}].$$ 0 You have $$\bigcup_{n=2}^\infty \left[ \frac 1n, 1- \frac1n \right] = (0,1)$$so I doubt it is true. When you find a counterexample, there is no need to start writing a proof. 0 If$g$and$h$are continuous on$[a,b]$, then they are uniformly continuous, hence obtains minimum and maximum. Hence you can get a lower bound and upper bound for$y$. If they are not continuous on$[a,b]$, the set is not necessary bounded. For example,$a=0,b=1, g(x)=0, h(0)=0 \quad\text{and}\quad h(x)=\frac{1}{x}, x\in (0,1].$0 In a non-compact space, sure. Just consider closed rays on the real line, e.g.$[n, \infty)$for$n$a natural number. In a compact space a family of closed sets that every finitely many have a non-empty intersection, will have a non-empty intersection as well. So the diameter is not needed in showing that the intersection is non-empty. 0 In order to show that a set$S$is bounded in$\mathbb{R}^n$, you must show that it is a subset of a "ball" of some size. A ball$B(a,x)$, where$a\in \mathbb{R}^n$and$x$is a real number is defined as the set of points that are distance less than$x$away from$a$.$B(0,x)=\{a\in \mathbb{R^n} : d(0,a)<x\}$In your example, you are working in$R^2$and ... 2 The intuition of a "french railway metric" alluded to in Dan's answer is very helpful. But one should avoid to be drawn into chasing cases. Given three points$x$,$y$,$z\in{\mathbb R}^2$denote by$M$the union of the three segments$[x,P]$,$[y,P]$,$[z,P]$. By definition, for any two points$u$,$v\in M$the distance$d(u,v)$is the length of the ... 0 Either method works. They essentially the same homotopy, just different expressions: The angle between$\vec{n}_t$and$\vec{n}_1$is the angle in the second homotopy. They fix the endpoint$p$because all planes$H_t$pass$p$and$q$. Note: You may need to pay attention to the orientation of the paths$C_1$and$C_2$, as the above homotopy may send ... 2 That follows immediately from the definition of convergence of a sequence. Let$X$be any topological space,$\sigma=\langle x_k:k\in\Bbb N\rangle$a sequence in$X$, and$x\in X$. Then by definition$\sigma$converges to to$x$if and only if for each nbhd$U$of$x$there is an$n_U\in\Bbb N$such that$x_k\in U$for all$k\ge n_U$. In your case ... 0 Let$A$a set. We want to prove that$A$is open. If$A$is infinite you are done. Otherwise,$X-A$is infinite and you can divide it in two infinite disjoint subsets$U_1$and$U_2$. Then,$U_1\cup A$and$U_1\cup A$are infinite too, so $$A = (U_1\cup A)\cap (U_2 \cup A)$$ is open. 0 As you already pointed out, sequences of the type$\left\{\left(-\dfrac{1} {n},y\right)\right\}$are not equivalent to$\left\{\left(\dfrac{1} {n},y\right)\right\}$, for$|y|\leq1$, so each one will converge to a different point in the complete space. If we call this points$(0^-,y)$and$(0^+,y)$their distances will be given by$d=2(1-|y|)$. So the space ... 0 Okay, now that I know what the problem is, I can tell you it's impossible. There is no way to link the torus with the line by continuous deformation. The concept of linking number is relevant here. All the curves on the original torus have zero linking number with the line, but in the linked picture, the core curve of the torus has linking number$\pm 1$... 1 Hints: 1) It suffices to show every singleton set$\{x\}$is open. 2) To see that that is the case, can you exhibit$\{x\}$as an intersection of two infinite subsets of$X$? 1 Let$S^0 = \{1,-1\}$. Then$S^0 \times I = \{1,-1\}\times I = \{1\} \times I \cup \{-1\} \times I$is basically a disjoint union of two copies of the unit interval$I$. Hence a homotopy$H$between two maps$f,g \colon S^0 \to X$"consists" of two paths in$X$, one,$t\mapsto H(1,t)$, connecting$f(1)$and$g(1)$, the other,$t \mapsto H(-1,t)$, connecting ... 1 A complex submanifold does not disconnect the complex space since the real codimension is$\ge 2$. The analytic approach: Say inside$\mathbb{C}^n$we have a subvariety$f=0$. Take$P$,$Q$point with$f(P) \ne 0$,$f(Q) \ne 0$. Consider a complex affine line$L$through$P$,$Q$( looks like the complex plane). The function$f$restricted to$L$is not ... 2 I'm assuming that you mean$T$is the smallest topology containing these sets -$T$itself is not for example closed under unions. Here is a thought for 2: How many open sets in your topology include the point$0$? What might this imply about an open cover in this topology? 5 Larson’s example in the paper cited by Tomek Kania and its verification are simple enough to be worth giving here (in very slightly modified form) for easy reference. Let$\tau$be the cofinite topology on an uncountable set$X$, let$\tau_0\subseteq\tau$be a$T_0$topology on$X$, and let$\tau_0^*=\tau_0\setminus\{\varnothing\}$. For each ... 1 It suffices to check that inverse images of$f$of basic open sets$U \times V$of$Y \times Z$(so$U \subset Y$open and$V \subset Z$open) are open. Then note that$f^{-1}[U \times V] = f_1^{-1}[U] \cap f_2^{-1}[V]$and proceed from that. For the other direction,$\pi_Y \circ f = f_1$, where$\pi_Y$is the projection onto$Y$. Likewise for the other ... 0 If$x \in C$, we have that$x \notin K$, so there is an$r_x > 0$such that$B(x, r_x) \cap K = \emptyset$(as$K$is closed). The set$\{B(x, \frac{r_x}{2}): x \in C \}$forms an open cover of$C$, so finitely many of them cover$C$, as$C$is compact, say$B(x_1, {r_{x_1} \over 2}),\ldots, B(x_N, {r_{x_N} \over 2})$cover$C$, and let$r ...

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Another way to get negative answers: a topological group does not have the fixed point property (every map from $X$ to itself has a fixed point), as a translation $x \rightarrow a\ast x$ with $a \neq 1$ does not have a fixed point. But there are homogeneous compact metric space with the fixed point property, e.g. the Hilbert cube $[0,1]^\mathbb{N}$, see ...

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I asked a related property: are there non-homeomorphic $X$ and $Y$ such that there is a continuous bijection from $X$ to $Y$ and from $Y$ to $X$ as well? And this mathoverflow answers shows that we can have such examples. They're a bit harder to come up with, I think.

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If you have a product $F^\mathbb{N}$ of spaces, then this product is the limit of an inverse system: let $X_n = F^n$ for all $n$, and the connecting functions $f_{n,m}: X_n \rightarrow X_m$, for $m \le n$ are just the projections onto the first $m$ coordinates. Then the limit of the inverse sequence we get this way is just $F^\mathbb{N}$. And the latter ...

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Take $[0,1]$ and $(0,1)$ with their standard topologies in $\Bbb R$. Clearly they are not homeomorphic. But $[0,1]$ is homeomorphic to $[\frac14,\frac13]$ and $(0,1)$ is homeomorphic to itself.

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No, consider $A=[0,1]$, $B=[0,1]\cup [2,3]$, $f:[0,1] \to [0,1]\cup [2,3]$ where $$f(x) = x$$ and $g(x): [0,1]\cup [2,3]\to [0,1]$ with $$g(x) = \frac{x}{6}$$

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In terms of $\theta$ the Riemannian metric on the half-hyperbola $H$ is given by $$ds=|f'(\theta)|\>d\theta=\sqrt{\sinh^2\theta +\cosh^2\theta}\ \ d\theta\ .$$ Now the variable $\theta$, where $-\infty<\theta<\infty$, is a global coordinate on $H$. Therefore the map \sigma:\quad H\to{\mathbb R},\qquad f(\theta)\mapsto\int_0^\theta ...

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It's a tricky point to get across, primarily because there is a deep mathematical theorem in the background, the Jordan Brouwer separation theorem. Although that wikipedia link uses the words "interior" and "exterior", those words have other meanings which causes ambiguity, as some of the comments point out. A better terminology would be one that is just as ...

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Here is some intuition as to why $D^n/S^{n-1} \simeq S^n$ is true. Observe that the boundary of $D^n$ is homeomorphic to $S^{n-1}$, so what this relationship is expressing is that if you glue the boundary of an $n$-dimensional ball to a point, what you get is the $n$-dimensional sphere. This can be nicely visualized in the case of $n=1$ and $n=2$. In the ...

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