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Assume for a contradiction that $x \in X$ is a cluster point of the $x_n$. Since the $U_{\alpha}$ cover $X$, there is some $\alpha$ such that $x \in U_{\alpha}$, and since $U_{\alpha}$ is open, for some $\epsilon > 0$, $B(x, \epsilon) \subseteq U_{\alpha}$. Now choose $N$ such that $1/N < \epsilon/2$. As $x$ is a cluster point of the $x_n$, $0 < ... 1 Here is a counterexample to the following claim: Let$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$be the unit ... 0 Since$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. 0 HINT: Show, if you’ve not already done so, that the projection map$\pi:X\times\Bbb R\to X$is continuous and open. (This is true of all projection maps.) Its restriction to$G(f)$is then easily seen to be continuous and open, and it’s not hard to check that this restriction is a bijection and hence a homeomorphism. 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 ... 3 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$X\cap B=\emptyset\RightarrowX\cup B$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. 1 Yes, whenever you have a norm$\|\cdot\|$on some space, it automatically gives you a metric$d(f,g) = \|f-g\|$. (One says that the norm induces the metric.) So, the norm$\|f\|=\int_0^1 |f(x)|\,dx$induces the metric$d(f,g)=\int_0^1 |f(x)-g(x)|\,dx$the norm$\|f\|=\sup_{0\le x\le1} |f(x)| $induces the metric$d(f,g)=\sup_{0\le x\le 1}|f(x)-g(x)|$, ... 1$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 ... 1 To “prove” this, you need convince yourself, that (i) every closed subset of a complete metric space is automatically complete with respect to the metric; (ii) the reals under the eukl. norm$(\mathbf{R},|\cdot|)$is a complete metric space; and (iii)$[a,b]\subseteq\mathbf{R}$is closed in this topology. Statement (i) is an easy exercise (take a ... 0 Hint: It is a well known result that given a Banach space$X$, and a (topological) subspace$F \subset X$, then$F$is closed if and only if$F$is complete. Then you should be able to conclude now, since$\Bbb R$is complete. Is$[-1,1]$closed in$\Bbb R$? 1 Let$x, y\in X$. Then if$d(f(x), f(y)) = 0$,$f(x) = f(y)$. Since$f$is a bijection,$x = y$. The rest of the properties are straightforward. 1 For example:$d_f(x,y) = 0$iff$d(f(x),f(y)) = 0$iff$f(x)= f(y)$iff$x=y$. Non negativity follows because$d$is non negative. For the triangle inequality:$d_f(x,y) = d(f(x),f(y)) \le d(f(x),f(z)) + d(f(z),f(y))= d_f(x,z)+d_f(z,y)$. 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 ... 2 I found a lot of stuff about the subject, so this answer will be rewritten too. It seems the following. We can answer your question positively via the following Proposition 1. Nevertheless, I am still thinking about another characterization, which will tell us more about a structure of the space$X$. We shall need the following definitions. A subset$A$... 2 First, note that a separation on a subspace of a topological space consists of two non-empty, disjoint sets that are open relative to the subspace. Here, you need two non-empty, disjoint sets that are open in$A \cup B $. It is an equivalent problem to prove the existence of two non-empty disjoint sets that are closed relative to$A \cup B. The sets ... 2 \begin{align*} U_1 &= \{x : \mathop{\text{dist}}(x,A) < \mathop{\text{dist}}(x,B) \} \\ U_2 &= \{x : \mathop{\text{dist}}(x,A) > \mathop{\text{dist}}(x,B) \} \end{align*} (The use of this kind of distance trick is suggested by the fact that the result is not true in general topological spaces; a nice counterexample is the co-finite topology on ... 1 The metric you described is the standard metric on the projective space: in the real case it can be visualized as the angle between lines (thinking of the elements as lines). It arises as the quotient of the spherical metric onS^n$by the group of isometries$\{x\mapsto \alpha x, \ |\alpha|=1\}$where$\alpha$belongs to the ground field,$\mathbb{R}$or ... 0 What you need to show is that if you pick an$x$that lies in the open ball around$c$, then it must also lie in the open ball around$a$. You know an upper bound to both d$(a,b)$, d$(x,c)$and d$(c,b)$(make sure you know why and how). So, write out d$(x,a)$, make use of the triangle inequality a couple of times, and conclude that the following holds: ... 1 Well this is not really a norm (because$X$is not even an abelian group) nevertheless$X$can be always seen as a metric space with the following distance : $$d(f,g):=\sum_{n=1}^{\infty}\frac{|f(n)-g(n)|}{2^n}$$ I think now of$X$with the distance written above. Define a sequence$(f_k)$: $$f_k(n):=n\text{ if } n\leq k\text{ and } 0 \text{ if } ... 0 It seems the following. I want to find an open set in X which is dense-in-itself. Is this possible? It is not always possible. Let \{a_n\} be an enumeration of all rational points of the unit segment [0,1]. Let$$X=\{(1/n, a_n):n\in\Bbb N\}\cup (\{0\}\times ([0,1]\setminus\Bbb Q)) \subset\Bbb R^2.$$Then X is an uncountable zero-dimensional ... 0 One way to do this, is the following: Prove that d: M \times M \to \Bbb R is a continuous function on M \times M. To do this, you can show the following Lemma: Lemma. Let (M, d) be a metric space. Then for all x,y,u,v \in M the inequality$$\vert d(x,y) - d(u,v) \vert \leq d(x,u) + d(y,v) $$is fullfilled. Now let \{ (x_n, y_n) \}_{n=1}^\infty ... 0 This is true. Check at the end of the Wikipedia page on proper maps for a proof: 2 Here is a simple counter-example which proves that X is not necessarily compact. Let X=[0,1) which is not closed then not compact. and f(x)=x(1-x) for x in X. Then the range is exactly [0,1] and f is continuous. EDIT. Topologies are usual ones on X and on the range. 6 No, it doesn't have to be. Counterexample: endow [0,1] with the discrete topology, \mathcal T_{dis}. Call the standard topology on [0,1] \mathcal T_{Euc}. Now consider the identity function:$$\mathrm{id}:([0,1],\mathcal T_{dis})\longrightarrow ([0,1],\mathcal T_{Euc}) \atop \qquad x\mapsto x$$Then f^{-1}(\{y\})=\{y\} for all y\in[0,1] yet ... 3 You are almost done: Assume (c) does not hold. Then there is \epsilon >0 such that: For each n, there is E_n so that diam (E_n) < \frac 1n and diam(F(E_n)) \ge \epsilon. Then there is x_n, y_n \in E_n so that d'(f(x_n), f(y_n)) \ge \epsilon/2. 2 Note that$$ X \setminus A = \bigcup_{x \in X \setminus A} \{ x \}. $$Therefore, by de Morgan’s Laws,$$ A = X ~ \Bigg\backslash \bigcup_{x \in X \setminus A} \{ x \} = \bigcap_{x \in X \setminus A} X \setminus \{ x \}, $$which makes A an intersection of open subsets of X because in a metric space, points are closed. It follows immediately that ... 0 The function d_1 fails positivity on \Bbb R as d_1(0,1)=-1<0 and d_1(x,x)=0 for all x\in\Bbb R. It also fails symmetry as d_1(1,0)=1\neq-1=d_1(0,1). In particular d_1 is not a metric. 1 Hints Please write down the definition of a metric. Which of the axioms can you find counter-examples for using this d_1? 2 Note that this is a special case of the general theorem of means. If a, b are non-negative and 0 < x, y < 1 with x + y = 1 then a^{x}b^{y} \leq ax + by. Equality occurs if a = b. Here in the current question a = t, b = 1, x = \theta, y = 1 - \theta. An easy proof of the general theorem mentioned above is based on Mean Value Theorem. ... 2 If you demand \delta<\frac12|x_0|, then the denominator is >\frac12|x_0|^2. I you additionally demand \delta<\frac12|x_0|^2\epsilon, you are done. 0 You may want to see the answers for this question, which answer yours, Extending a function by continuity from a dense subset of a space. I built the proof myself based on Srivatsan's answer for that question. If anybody still needs it, here it goes: Theorem If X and Y are metric spaces and f:S \to Y is uniformly continuous with S dense in X, ... 1 If I understand you correctly, the objects you are considering are finite sets and your dissimilarity function is defined as d(A, B) = \frac{|A \ominus B|}{|A \cup B|}, where \ominus denotes the symmetric difference. Consider the Venn diagram for three sets A, B, and C. Then the triangle inequality boils down to$$ \frac{a+b+e+f}{a+b+d+e+f+g} + ... 3 Your example is fine, but your argument is not:$\Bbb R$with the usual topology, for instance, is second countable but has uncountably many open sets. Let$\mathscr{B}$be a base for the discrete topology on$\Bbb R$. For each$x\in\Bbb R$the set$\{x\}$is open, so for each$x\in\Bbb R$there is a$B_x\in\mathscr{B}$such that$x\in B_x\subseteq\{x\}$. ... 2 If we take$g:\theta\to t^{\theta}$we are just stating that the graphics of$g$on the interval$(0,1)$lies below the line through$(0,1)$and$(1,g(1))$. However, that is trivial, since$g(\theta)$is a convex function due to: $$g''(\theta) = t^{\theta}\log^2 t \geq 0.$$ 4 For$t > 0$, write $$t^\theta = t^\theta\cdot 1^{1-\theta}.$$ Taking the logarithm of both sides, what you need to show becomes $$\theta\log t + (1-\theta) \log 1 \leqslant \log (\theta\cdot t + (1-\theta)\cdot 1),$$ which follows from the concavity of$\log$. 3 I don't know if this is correct, but I gave it my best shot: Lemma: A continuous function$f:X \to \mathbb{R}$is good if and only if it is compactly supported. Proof: Suppose that$f$is compactly supported, and let$C = $supp$f$denote the support of$f$. Then for any continuous map$g:X \to \mathbb{R}$, the set$\{x : f(x)g(x)=1\}$is a closed subset of ... 0 The problem is: Let$X$be a finite topological space obtained from a metric, prove$X$is discrete. You just have to prove that singleton subsets of$X$are open. This is because if the singleton subsets are open then the union of any family of singleton sets is open, but of course every subset$A$can be seen as$\bigcup\limits_{a\in A}\{a\}$. To prove ... 0 Counter example could be the discrete metric space(on$\mathbb{R})$.$d(x,y)=1$, if$x \ne y$, else$d(x,y)=0$. To show that the statement stands: For example, take$ϵ=\frac13$. Then the ball will only contain the point itself, so it will be a subset of our set. Every subset will be closed because the complement of an open set is closed. 2 Yes. You have used the axiom of countable choice to choose$x_n$from each$B_{\frac1n}(x)\cap A$. To wit, in Cohen's first model there exists a dense subset$D$of$\Bbb R$which is Dedekind-finite. Namely$D$has no countably infinite subset. It is not hard to see, if so, that any$x\in\Bbb R\setminus D$is in the closure of$D$, but no sequence in$D$... 0 Suppose$(x_n)$be a sequence in$f^-$$^1$$[F]$converging to x∈$R$since$f$is continuous on$R$that means$f$is continuous on each point of$R$By$Sequential-Criterian$of continuity$f(x_n)$converges to$f(x)$.$F$is closed and$f(x_n)$∈$F$implies that$f(x)$∈$F$. Finally we have$x$∈$f^-$$^1$$[F]$, Hence$f^-$$^1$$[F]$is closed. 1 I will write this in a slightly different way, but the idea is similar to copper.hat's proof. Question: Let$A$be a closed set and$K$be a compact set such that$A$and$K$are disjoint. Define the distance $$d(A,K) = \inf \{d(a,k): a\in A, k\in K\}$$ Prove there exists$r > 0$such that$A$and$\bigcup_{k\in K} \overline{D}(k;r)$are disjoint. ... 0 Take an open ball$B$. Since$A$is nowhere dense,$B$cannot be contained in$\overline{A}$. Pick$b \in B, b \notin \overline{A}$. Since$\overline{A}^c$is open, there is a ball$B'$containing$b$that doesn't intersect$A$. Let$U:=B \cap B'$. Now, pick an open ball$B_1 \subset U$and we are done. 2 That step is not legitimate. Consider the set$A=\left\{\frac1n:n\in\Bbb Z^+\right\}$in$\Bbb R$. For each$\epsilon>0$we have$B(0,\epsilon)\cap A\ne\varnothing$, but we can’t take the limit as$\epsilon\to 0^+$to conclude that$\{0\}\cap A\ne\varnothing$: this is clearly false. Try this instead. Suppose that every open ball contained in ... 2 Your proof is good, here is another that avoids explicit sequences. Let$\phi(x) = \inf_{w \in V^c} d(x,w)$. We see that$\phi(x) > 0$iff$x \in V$since$V$is open.$\phi$'measures' the distance from$x$to the complement of$V$. Since$d(x,v) \le d(x,y)+d(y,v)$, we see that$\phi(x) \le d(x,y) + \phi(y)$, and swapping$x,y$shows that ... 1 One direction is easy: if$\overline{X}$is compact, then$\overline{X}$is totally bounded, so that$X \subset \overline{X}$is totally bounded. For the other direction, suppose$X$is totally bounded. Then$\overline{X}$is clearly complete, so all we need to do is to prove that$\overline{X}$is totally bounded. Pick some$\varepsilon>0\$. Choose ...