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1

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 $X\cap B=\emptyset$ $\Rightarrow$ $X\cup B$ 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.


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 on $S^n$ 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 ...



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