# Tag Info

7

You have pretty much got it though there are some minor errors in your proof. For each $n$ we consider the open cover $C_n=\{B(x, 1/n):\ x\in X\}$ of $X$. By compactness, there is a finite subset $X_n$ of $X$ such that $B_n:=\{B(x, 1/n):\ x\in X_n\}$ covers $X$. (Here you made a mistake in your proof by implicitly assuming that $X_n$ can be chosen to have ...

5

This is a global condition - that is, it is both necessary and sufficient to have your condition be true for all $x,\delta$. You need: (Condition 1): Given any $x\neq y$ and any $\epsilon>0$ that there is some $z$ so that $d(y,z)<\epsilon$ and$d(x,z)<d(x,y)$. That is, every neighborhood of $y\neq x$ has a point closer to $x$ than $y$ is. ...

5

I understand the bi-invariance as $$d(ax,ay)=d(x,y)=d(xa,ya)$$ for any $a,x,y\in G$. Then $$d(x,y)=d(1,x^{-1}y)=d(y^{-1},x^{-1})=d(x^{-1},y^{-1}).$$ The last step is the symmetry of $d$.

5

It is not a matter of "thinking", "considering" or "debating". Your professor perhaps has given a definition of open ball. Or, at least, he must have assumed some definition. That definition should specify if the radius of the ball must be a positive number or null radii are allowed. From my experience, most books that include a definition of open ball say ...

4

Start with the fact that for any homeomorphism of topological spaces $f:X \to Y$ and for any $x \in X$ and $y=f(x) \in Y$, the restricted function $f : X-\{x\} \to Y-\{y\}$ is a homeomorphism using the subspace topologies. So for your problem it suffices to show that if $X=Y=B_1[0]$ and $x \in S^n$ and $y \in B_1[0]-S^n$ then $B_1[0] - \{x\}$ is not ...

4

The triangle inequality implies that $|d(y,x) - d(y',x)| \leq d(y,y').$ It follows that $d(-,x) : X \to \mathbb{R}$ is continuous (in fact, Lipschitz). Hence, the preimage under the open subset $]r,\infty[ \subseteq \mathbb{R}$ is open, and this is your set.

3

Let $y\in X$ such that $d(y,x)>r$, and choose $\epsilon$ such that $0 < \epsilon < d(x,y) - r$. We claim that $$B(y,\epsilon):= \{z\in X|d(y,z)<\epsilon\} \subseteq \{y\in X|d(y,x)>r\},$$ so the open ball $B(y,\epsilon)$ is contained in any arbitrary point $y$ in the given subset of $X$. Let $z \in B(y,\epsilon)$, then $d(y,z)<\epsilon ... 3 Let$U = (-1,1)$,$V_x = (-1,1)$for$x = 0$and$V_x = (-\frac{1}{2}, \frac{1}{2})$otherwise. There is obviously no$\varepsilon$-Ball around$(0, \frac{3}{4})$. 3 First of all, Vim's comment is absolutely right: you might need more than$n$-many balls of radius${1\over n}$to cover your space. But of course this doesn't affect the argument, which only needs there to be finitely many. So let's let$x_{n, i}$($i\le k_n$) be the corresponding centers. Hint: If$U$is open and$x\in U$, then there is some$s$such that ... 3 You need to redefine your$\Bbb B$. For each$n\in \Bbb N^+$the corresponding open cover$B(x,1/n),x\in X$admits a finite subcover $$B_n:=\{B(x_{ni},1/n),i=1,2,\cdots,k_n\}$$ due to compactness of$X$. Now you want to show$\Bbb B:=\cup B_n$is a desired basis. Countableness has been shown in your attempt, it only suffices to show$\Bbb B$is a ... 3 One possibility is to use the Invariance of Domain theorem (which in turn relies on non-trivial topology). Let$h\colon B_1\to B_1$be a homeomorphism. Here$B_1$is the closed unit ball in$\mathbb{R}^n$. Let$\mathring{B}_1$denote the open unit ball. The restriction of$h$to$\mathring{B}_1$is an injective continuous map of a domain of$\mathbb{R}^n$... 3 Can you use continuity and the Intermediate Value Theorem for a Connected domain to show that the function $$\phi(z)-\phi(-z)=0$$ for some$z$? If so, you are done. 2 The open character of A and B can be deduced from the definition of open set. Let$a\in A$. Supposing$b\in B$^,$(a,b)\in A\times B$. As$A\times B$is open, there exists$B((a,b),r)\subset A\times B$. Then if$a'\in B(a,r)$,$d_X(a,a')=d_X(a,a')+d_Y(b,b')=d_1((a,b),(a',b))<r$, so$(a,b)\in A\times B$, and thus$a'\in A$. That is, A is open. An analogous ... 2 Not sure why you use$a$both for elements in$\mathbb R$and in$X$. Note by triangle inequality, $$d(x, a)\le d(x, y)+ d(y, a)\Rightarrow d(a, x) - d(a, y)\le d(x, y)$$ Interchanging the role of$x, y$and use$f(x) = d(a, x)$we have $$\tag{1}|f(x) - f(y)|\le d(x, y).$$ This inequality is sufficient for us to condlude that$f$is continuous. Let$U$... 2 In a metric space$(X,d)$, you define the metric interval between$x,y\in X$as $$I(x,y) = \{z \in X \mid d(x,y) = d(x,z) + d(z,y)\}$$ Note that this metric interval is always closed (think sequences and continuity of$d$). With this notion, you can start to define and study things like convex and hyperconvex metric spaces, etc, despite not having an ... 2 The theorem in your book says that "if two points in$Y$are to close, then their preimages are to close" (remember: closeness is measurable by mean of open sets in general topology). It has a good corollary:$f$is continuous iff$f^{-1}(V)$is closed for every closed$V$. Now, i'll suppose you are working with the usual topology. In your case, note that ... 2 Hint: Let$X$be the reals in$(0,1)$with the usual metric, and$Y$the reals with the usual metric. Let$f(x)=\tan(\pi x/2)$. 2 Equivalent metrics gives the same topology, so we can show that the metrics are equivalent, I'll replace$d(x_1,y_1)=x$and$d(x_2,y_2)=y$and show that they are equivalent. Remember 2 metrics are equivalent if$c d_2(x,y)\leq d_1(x,y)\leq C d_1(x,y)$for some$c,C\in\mathbb{R}$always holds for all$x,y\in M$. We will show that i and iii are eqivalent and ... 1 Let$f$be a function in$A$.$f$belonging to the interior of$A$means that there is a "small ball" in$A$wrapping it. So to disprove it you need to find a way to show that this kind of "small ball"s does not exist. The simplest and straightest way is to find a sequence$(f_n)_{n\in\mathbb{N}} \notin A$that converges to$f$in this metric. For example ... 1 Another approach. Let $$U=\{f\in C[0,1]:\mbox{ there is K such that}|f(x)-f(x_0)|\le K|x-x_0|\,\,\forall x\in[0,1]\}$$ Note that$A\subset U$. On the other hand, if$f,g\in U$and$a$is scalar:$\begin{eqnarray}|(f+ag)(x)-(f+ag)(x_0)|&=&|f(x)+ag(x)-f(x_0)-ag(x_0)|\\ &\le&|f(x)-f(x_0)|+|ag(x)-ag(x_0)|\\ ...

1

(some changes in order to improve precision) About your last interrogation. Any normed vector space defines naturally a metric space by the relationship $d(x,y)=\|x-y\|_p$. Thus, indeed, any $\|\cdot\|_p$ norm induces naturally an $\ell_p$ "metrics" (synonym : "$\ell_p$ distance"). A point of vocabulary about the words "metrics" vs. "distance". "Metrics" ...

1

Suppose $x_n$ is a sequence in $Z(f)$ such that $x_n\to x\in X.$ We want to show $x \in Z(f).$ By continuity, $f(x_n) \to f(x).$ But $f(x_n) = 0$ for all $n,$ so …

1

I think you need a somewhat different proof. For each $n$ consider $\{B(x,1/n):x\in X\}$. Refine it to a finite cover using compactness. The union of those finite covers for all $n$ is a countable set of balls. Let the $x_i$ be the centers of those balls. For every $x$ and every $n$, $x$ has to be in the finite cover corresponding to $n$, and thus there ...

1

Hint: your space is complete, so Cauchy $\implies$ convergent. Now, what happens with the pointwise limit?

1

For the first question, note that $\cos(x)$ is positive and decreasing for $x \in[0, \frac{\pi}{2}]$. So $f_n(x)$ is also positive and decreasing on $[0,1]$, thus we have $$d(1,f_n) = \sup_{x \in[0,1]} 1 - \cos \left( \frac{x}{n} \right) = 1 - \cos \left( \frac{1}{n} \right)$$ So you are left to show that $\cos \left( \frac{1}{n} \right) \to 1$ as $n \to ... 1 Hint for an explicit bound in (1): $$\forall x\in\Bbb R:\ 1-\frac{x^2}2\le\cos x\le 1.$$ (2): Yes, because$|b\cos(x/n)-b| = |b|\,|\cos(x/n)-1|$. (3): Yes. 1 Your proof is good, but too lengthy. Suppose$E$is not bounded. For each positive integer$n$, there is$x_n\in E$such that$\|x_n\|>n$. I claim that$S=\{x_n:n>0\}$is infinite. If it is finite, then$M=\max\{\|x_n\|:n>0\}$exists; but if$n$is the least integer greater than$M$, we have$\|x_n\|>n>M$: a contradiction. The set$S$cannot ... 1 A necessary and sufficient condition on a metric space$(X,d)$for$d$to come from some pre-metric is that$(X,d)$be isometric to a subspace of the metric space$\mathbb{R}$with the ordinary distance. We may assume$X \ne \varnothing$. (When$X = \varnothing$, both conditions are true.) Let$f$be a pre-metric on$X$, and fix an "origin"$a \in X$. ... 1$A\times B$is open implies for every$(a,b)\in A\times B$, there exists a ball$B((a,b),r)\subset A\times B, r>0$.$(x,y)\in B((a,b),r)$i.e$d_X(x,a)+d_Y(y,b)<r$. Let$B(a,r)=\{x\in X:d_X(x,a)<r\}$, we have$B(a,r)\times b\subset B((a,b),r)\subset A\times B$. So$B(a,r)\subset A$and$A$is open. Same argument with$B$. 1 Since it is suggested that connectedness be explicitly used, I might phrase it like this: The image of the connected domain of the function$\mathbf z \mapsto \phi(\mathbf z) - \phi(-\mathbf z)$must be a connected subset of$\mathbb R$. If it is not everywhere$0$, then for any point$\mathbf z_0$where it is not$0$, the function changes signs as$\mathbf ...

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