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The function (call it $g$) is actually not continuous in general. Note that by definition $g(0) = 0$. For example let $X = \mathbb S^1$ with the subspace metric of $\mathbb R^2$ and $f : \mathbb S^1 \times \mathbb R \to \mathbb S^1$ be given by $$f(e^{i\theta}, t) = e^{i(\theta + t^2)}.$$ For all $s\neq 0$, let $t = \frac{\pi-s^2}{2s}$. Then for all $x = ... 1 Suppose$inf\{\rho(x_0,a), a\in A\}=0$. This implies that for every integer$n>0$, there exists$a_n\in A$such that$\rho(x_0,a_n)<1/n$. This implies that the sequence$a_n$converges to$x_0$thus$x_0\in A$since$A$is closed. Contradiction. done. 1 HINT:$\beta$is basically the "distance" from$x_0$to the set$A$. Unfortunately, sometimes we have$\beta=0$even though$x_0$isn't actually in$A$- for example, take$A=(0, 1)$and$\beta=0$. Then$x_0\not\in A$, but there are elements of$A$"arbitrarily close" to$x_0$, so$\beta$- the infimum (over$a\in A$) of the distances from$x_0$to$a$- is ... 1 Define$g:K\times K$by$g(x,y)=\left \| x-y \right \|$. As$g$is continuous, and$K\times K$is compact, there is an$(x_0,y_0)\in K\times K$such that$g(x_0,y_0)=\left \| x_0-y_0 \right \|$is a maximum. But then, since by hypothesis,$(f(x_0),f(y_0))\in K\times K$we have$\left \| \frac{f(x_0)-f(y_0)}{x_0-y_0} \right \|\leq 1\Rightarrow \left \| ...

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HINT: Show that for any $x\in\Bbb R$, $(x,x+1)\subseteq(\lfloor x\rfloor,\lfloor x\rfloor+2)$. For each $C\in\mathscr{C}$ it’s very easy to find an $A\in\mathscr{A}$ such that $C\subseteq A$; how? For $\mathscr{D}$ consider $x=\frac34$.

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@AsafKaragila Impressive. I would like to extend my understanding from two sides: If we extend $\bar{d}(x,y) = \min\{d(x,y),1\}$ from $\mathbb R^1$ to $\mathbb R^\omega$, then $\bar{d}(\mathbf x,\mathbf y) = \min\{d(x_\alpha,y_\alpha),1\}$, $\alpha\in\omega$. Can we interpret it as this example: $\bar{d}(\mathbf x,\mathbf y) = \min\{d(x_1,y_1),1\}$, ...

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I assume you are not familiar with the topological concept of continuity (that's the answer everyone thinks about). Now, obviously, $A$ is bounded so it suffices to prove that it is in fact closed, that is, its complement is open. Let $p\in\mathbb R^n\setminus A.$ We shall prove that $p$ is an interior point of $\mathbb R^n\setminus A.$ As $p\notin A$ ...

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A is open so every point in A, especially points of C have an open neighbourhood of radius rx contained A, so the distance to the boundary is larger than or equal to rx. Using the compactness of C we now can select a finite subcover and find a positive distance fro C to boundary of A.

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As $x \in C$, we know that $x \in A$ which is open. So there is some $r>0$ such that $B(x,r) \subseteq A$. All points in $B(x,r)$ are not in $\operatorname{Fr}(A)$ (as they are interior points of $A$), so $d(x,\operatorname{Fr}(A)) \ge r > 0$.

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Edit: The following answers a former version of the question, referring to a subset of $S^{n-1}$. The statement in the title, appearing again in the body of the question, is wrong. A subset $$A\subset S^{n-1}\subset {\mathbb R}^n$$ is compact iff it is closed. Consider, e.g., the set $$A:=\bigl\{x\in{\mathbb R}^n\>\bigm|\>\|x\|=1, \ x_n>0\bigr\}\ ... 1 Let the equivalence relation \rho be given by a\rho b\iff d(a,b)<\infty Now d(a,a)=0\forall a\in X thus \rho  is reflexive. Let a\rho b; then d(a,b)<\infty\implies d(b,a)<\infty \implies b\rho a\implies \rho is symmetric. Again let a\rho b, b\rho c then d(a,b)<\infty, d(b,c)<\infty,By triangle inequality we have ... 2 You are asked to prove that the relation on X given by x \sim y \Leftrightarrow d(x, y) \not = \infty is an equivalence relation. Concretely, that is: d(x, x) \not = \infty; d(x, y) \not = \infty means d(y, x) \not = \infty; if d(x, y) \not = \infty and d(y, z) \not = \infty, then d(x, z) \not = \infty. 4 Just note that every norm is continuous (because 1-lipschitz from the second triangle inequality |\,||x||-||y||\,|\leq||x-y||) and then A=||\cdot||^{-1}(\{1\}) is a closed set as a reciprocal image of the closed set \{1\} of \mathbb{R} by the continuous map ||\cdot||. 2 Consider T(x_1,x_2,\dots) = (x_1, x_2-x_1,x_3-x_2,\dots). Then A = T(l^{\infty}). Note that T may be inverted by the following operation (y_1, y_2, \dots) \to (y_1, y_1 + y_2, y_1+y_2 +y_3,\dots). From your argument, if y \in A, you must have |\sum_{i=0}^n y_i| \leq K for all n for some fixed K. Let B = \{y \in l^{\infty} : |\sum_{i=0}^n ... 1 \ell^1 is a metric space and its closure equals to the set of all elements that have distance 0 from A. On one side, any element g of$$ B:=\{f\in\ell^1:\,\,\forall\,\,n\,\,|f(n)|\leq \frac{1}{2^n}\} $$has distance 0 from A, because it either in A or a limit of elements in A. On the other side, \ell^1\setminus B is open (being equal to ... 1 The uniform norm is bounded by the D norm but not vice versa. Consider a sequence of sine waves scaled so that the frequency increases very rapidly and the amplitude converges slowly to 0. 1 A is not necessarily closed, but that does not matter: any sequence has a cluster point in the compact metric space E (which is all we need). Using your r we indeed get infinitely many p_i that are in B(p, \frac{r}{2}) so for some large enough index n we indeed have that B(p, \frac{1}{n}) \subseteq B(p, \frac{r}{2}). But we need (for a ... 0$$1+\sum_{n=1} p^{nk}=1+\sum_{n=1} (p^k)^{n}=\sum_{n=0}^m (p^k)^{n}=\frac{p^{k(m+1)}-1}{p^k-1}$$in your example you have p=6 and k=2 so we have$$\frac{36^{(m+1)}-1}{35}$$So what number can be subtract from this such that it gives a power of two number? -\frac{1}{35} works as we get ... 0 As you use the tag "metric spaces" we can assume we are in a metric space. But then the statement is quite trivial, as all sets of the form \{x\}, x \in X are closed, and we can write any set A as \cup \{\{x\}: x \in A \}. So nothing about openness and almost nothing of the metric is then used. It is true that every open set in a metric is the ... 0 For any set X, the map (x,y)\mapsto 0 is a pseudometric, as both sides of the triangle inequality would always be zero, but is only a metric if X consists of a single point, as otherwise there would be x\ne y with d(x,y)=0. More interesting, though, is the discrete metric:$$ d(x,y) = \begin{cases} 1,& x\ne y\\ 0,& x=y.\end{cases} $$A ... 0 As long as this satisfies the three requisite conditions to be a metric, (X,d) is a metric space. 1) Triangle Inequality: d(x,y)\le d(x,z)+d(z,y)\:\forall x,y,z\in X. True because 0 is the only element of the metric space. 2) Positive Definiteness: d(x,y)\ge0\:\forall x,y\in X, d(x,y)=0 if and only if x=y. This also holds, as you can verify ... 0 Yes, this is the same thing as taking the discrete metric of a space and considering only singleton subsets. 1 If X = \{0\} and you define d(x,y) = 0 for each x,y \in X, we have d(x,x) = 0 for each x \in X (this is the only case). d(x,y) > 0 for each x,y\in X, x \neq y (this is vacuously true). d(x,y) = d(y,x) for each x,y\in X (this is also vacuously true). d(x,z) \leq d(x,y) + d(y,z) for each x,y,z \in X (true because 0 \leq 0 + 0). ... 0 If (e_n)_n \notin \ell^1, show that A  is not even bounded in \ell^1 and thus not totally bounded. For the other direction, show that A  is compact by showing that it is sequentially compact. Indeed, for a sequence (x_n^{(m)})_n \in A , use the usual diagonal argument to construct a subsequence (x_n^{(m_k)})_n  such that the limit x_n =\lim_k ... 0 d(x,y) \leq 1 for all x,y. Now, you can easily see that d(0,n) \rightarrow 1 when n \rightarrow \infty, so diam(A) can't be less than one. Hence, diam(A) \geq 1, and since diam(A) = sup\{d(x,y) : x,y \in A\} \leq 1 (the supremum is the least upper bound), we have that diam(A)=1. 1 If I understand your question correctly, it is the following: Let \epsilon > 0 be given. For what \alpha, \beta \in \mathbb C does the following hold:$$ \left|\frac{\alpha+z_0}{\beta z_1+z_0}\right|<\epsilon \quad \text{ for all } z_0, z_1 \in \mathbb C \text{ with } |z_0|<\epsilon \text{ and } |z_1|<\epsilon \, . $$First let ... 0 1) If d(x,y)=a<1 then \overline{d}=\min\{a,1\}=a. 2) If d(x,y)=b\geq 1 then \overline{d}=\min\{b,1\}=1. 3) For any value of d(x,y), we always have \overline{d}\leq 1. 0 The idea is that metric is concerned with things "very close" rather than "very far". So if you define \bar d as you do, you don't lose a whole lot of information. This is like looking at the horizon. You will have the same information about what you can see even if you treat everything past the horizon as the same distance from you. So if I look ... 0 Everything you said is correct. The first two follow from the definition of minimum and the last one is a consequence of the definition of minimum. 1 Pick \varepsilon > 0. For each x \in X pick a neighbourhood U_x such that \mathcal{F} is equicontinuous at x for \frac{\varepsilon}{2}. Let \delta be the Lebesgue number for the cover \{U_x: x \in X\}. This \delta works (check this!). 2 The case d = 0 is uninteresting, since \mathbb{R}^0 = \{0\} has diameter 0. So we assume d \geqslant 1. Furthermore, if L is a straight line in \mathbb{R}^d, then the orthogonal projection \pi to L is Lipschitz-continuous with Lipschitz constant 1, hence \pi(B) is a connected subset of L \cong \mathbb{R} (where \cong means an ... 2 The answer to your question is included in the third line of the proof of the theorem you are interested in! By a classical result of Aharoni (see Theorem 7.11, p. 176 in 1), we know that there is a 3-Lipschitz-homeomorphism between C(K) and some subset of c_0. This proof can be summarized in four steps as follows: By Sobczyk's theorem, any ... 0 That proof starts with the sentence that c_0 is 2-complemented in C(K). And indeed c_0 is isometrically embedded into C(K). This holds as K is an infinite compact metric space, so has a convergent sequence (with all different terms) x_n with limit x (unequal to all x_n). Call S = \{x_n: n \in \mathbb{N}\} \cup \{x\}. Then C(S) embeds ... 1 Define r(x) = \min \{d(x,y): y \in X, y \neq x\}. This is a minimum of finitely many strictly positive numbers (as all d(x,y) > 0 when x \neq y). So r(x) > 0. Suppose y \in B(x,r(x)) and y \neq x. Then by definition of being in the ball d(x,y) < r(x) but r(x) \le d(x,y) by definition of r(x). Contradiction. So B(x, r(x)) = \{x\} ... 2 The notion of a metric is supposed to capture the main features of the everyday idea of distance between two points. The most important parts of this idea are that a point is at zero distance from itself; distinct points are a positive distance apart; the distance from a point x to a point y is the same as the distance from y to x; and the ... 0 First, in a complete metric space, being totally bounded is the same as being precompact. There is a simple characterization of precompact (totally bounded) sets of \ell^1. A subset A\subset \ell^1 is precompact if and only if For all n we have \sup_{f\in A} |f(n)| < \infty, and \sup_{f\in A} \sum\limits_{n=m}^\infty |f(n)| \to 0 ... 2 Say U is an open set. If U is empty or if it is the whole space X then U is also closed, so we are done. Now assume that U and its complement U^c=X\setminus U are both non-empty. Let A_n = \bigcup_{x\in U^c} B(x,\dfrac1n). Notice that A_n is open and its complement A_n^c (which is closed) is contained in U. To finish off the problem ... 2 Even more is true. Every copy of c_0 in C(K) for K compact, metric is complemented by a projection of norm at most 2. Indeed, C(K) is in this case separable (as K is second-countable we may use the Stone–Weierstrass theorem to get the claim) and then we may apply Sobczyk's theorem. 0 Let t_n \in K be a convergent sequence of distinct points with limit t_\infty also distinct from all t_n, consider C_0(K)=\lbrace f\in C(K): f(t_\infty)=0\rbrace and consider P:C_0(K)\to c_0, f\mapsto (f(t_n))_{n\in\mathbb N}. Conversely, choose peak functions \varphi_n \in C(K) with disjoint supports and \varphi_n(t_k)=\delta_{n,k} and ... 1 HINTS:$$\between\bullet\quad\bullet$$1 Hint: Consider the sets$$ U_1 = \{ x \in M \, | d(x, E_2) > d(x, E_1) \}, \,\,\, U_2 = \{ x \in M \, | d(x, E_1) > d(x, E_2) \}. $$0 HINT (a) Consider intersecting an annulus by a small and narrow rectangle in the plane. Can you construct a similar one for (b)? 1 Write Y = U \cup V with U,V open and U \cap V = \emptyset. Then f^{-1}(U) \cup f^{-1}(V) = X with f^{-1}(U), f^{-1}(V) open and f^{-1}(U) \cap f^{-1}(V) = \emptyset. Thus, f^{-1}(U) = X or f^{-1}(V) = X. If f^{-1}(U) = X, use the surjectivity of f to deduce that U = Y. Similarly for the other case. 1 When X is finite, you can treat \mathcal P as a subset of \Bbb R^X. Total variation is the norm there, and all the norms over finite-dimensional spaces are equivalent, so the induced metric is equivalent to the Eucledian one. Embedded \mathcal P is bounded and closed in Eucledian metric, hence compact. 1 Since you didn't state which definition of openness you're referring to, i'll use this one: A set A is open if for all a \in A there exists an open (bounded) ball B such that a \in B \subseteq A. Now, let B' = \{ y : d(x,y) > r\} and b \in B'. Let r' := d(x,b) -r > 0. Then we claim that \{y: d(b,y) < r'\} \subseteq B'. Take y such ... 0 Let (M, d) be a metric space; let x \in M. Then d(x, \cdot): M \to \Bbb{R} is continuous on M. If r \geq 0, then ]r, \infty[ is open in \Bbb{R}; hence the preimage of ]r, \infty[ under d(x, \cdot) is open in M. 1 If (x_n) is a Cauchy sequence, then there exists K\in\Bbb N such that |x_n-x_m|<1. Then it follows from the completeness of \Bbb R with the usual absolute value. 2 You've got two functions in play: f:I\to R^2 g : R^2 \to R, given by g(x) = \left| x \right| So what you're actually looking to do is show that g \circ f assumes a maximum value on I. But g \circ f is a function from I to R, and you already know this is the case for such functions so long as g \circ f is continuous. You know f is ... 0 If you can already show the analogous property of functions I\rightarrow\mathbb R, then the only step remaining is to note that the function g:\mathbb R^2\rightarrow\mathbb R taking vectors v to their norm |v| is continuous. This is easily proven by the triangle inequality. Then, we see that the map taking x to |f(x)| is simply the composition ... 0 For any r>0, we have$$(A-r)<d(a_n,c_n)\land d(a_n,b_n)<(B+r)\land d(b_n,c_n)<(C+r)$$for all but finitely many n. So for all but finitely many n we have$$B+C-A>(d(a_n,b_n)+d(b_n,c_n)-d(a_n,c_n))-3r\geq -3r. So $B+C-A>-3r$ for every positive $r.$ Therefore $B+C-A\geq 0.$

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