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This is completely equivalent to the normal notion of a metric. Specifically, a function $d(a,b)$ satisfies your axioms (1-4) iff the function $\log d(a,b)$ satisfies the usual definition of a metric. It is also easy to see that such a $d$ generates the same topology as the metric $\log d$ (since $\log$ is monotone and $\log x$ goes to $0$ as $x$ goes to ...

7

Eric Wofsey's comments are very much to the point: your notion of "metric" is just the thing you get when you apply the $\exp$ function to a standard metric. However, there are some things to be said about the specific example you have given. The metric you have defined (or, more precisely the metric $d(a,b)=\lvert \log a- \log b \rvert$) is an invariant ...

4

It is continuous. One way to see this is by using the fact that $x \mapsto x^2$ is continuous in $\mathbb{Q}$. Hence, the sets $A=\{x \in \mathbb{Q} \mid x^2 <2\}, B=\{x \in \mathbb{Q} \mid x^2 >2\}$ are open. Since the pre-images of open sets of your function $f$ are $A,B,\emptyset$ and $\mathbb{Q}$, it follows that $f$ is continuous.

3

Here is a some direction for the kind of proof I think you are looking for: $f: \mathbb{Q} \to \mathbb{R}$ is defined so that $f(x) = 1$ for $x^2 < 2$ and $f(x) = 0$ for $x^2 > 2$. We would like to show that $f$ is continuous. In other words, we need to show that $f$ is continuous at any point $a \in \mathbb{Q}$. Because $\sqrt{2} \not\in ... 2 You've got two functions in play:$f:I\to R^2g : 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$is ... 2 Even more is true. Every copy of$c_0$in$C(K)$for$K$compact, metric is complemented. Indeed,$C(K)$is in this case separable (as$K$is second-countable we may use the Stone–Weierstrass theorem to get the claim) and then we may apply Sobczyk's theorem. 2 Say$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$. To finish off the problem ... 2 This is not a mathematical term but a figure of speech. One says that A is an honest B to assert that A satisfies the definition of B. Usually the reason to emphasize honest is that A was informally called "B" earlier in the text. This is associated with a somewhat conversational style of writing. Specifically for distance functions, this word would ... 2 Not only are regular Lindelöf not necessarily second-countable, we can show that even stronger assumptions have no bearing on the weight of the topological space. (The weight$w(X)$of a topological space$X$is the minimum cardinality of a base for$X$, so that "$X$is second-countable" is equivalent to$w(X) \leq \aleph_0$.) Note that even compact ... 1$\Bbb R_\ell$(real numbers with lower limit topology) is a regular Lindelof space which is not second countable. See Munkres - Topology Chapter 4. 1 Let$C\subset Y$be closed in$Y$. We need to show that$f^{-1}(C)$is closed in$X$. $$f^{-1}(C)=\bigcup_{i=1}^nf^{-1}|_{F_i}(C)$$ Since each$f^{-1}|_{F_i}(C)$is closed. So,$f^{-1}(C)$is closed. 1 Hint: Consider the sets $$U_1 = \{ x \in M \, | d(x, E_2) > d(x, E_1) \}, \,\,\, U_2 = \{ x \in M \, | d(x, E_1) > d(x, E_2) \}.$$ 1 As$y \in f[F_n]$for all$n$, we have$x_n \in F_n$such that$f(x_n) = y$. As$X$(or$F_1$, which is enough) is compact, we know there is a convergent subsequence$n_1 < n_2 < n_3 < \ldots$such that$x_{n_k}$converges to some$x \in X$. This is the required subsequence. Pick any$m$. Then there is some$n_k$such that$m < n_k$, and so ... 1 Our distance function only takes on non-negative integer values. So if$d(A,B)\le \frac{3}{4}$, then$d(A,B)=0$. And it is easy to verify that the distance between two sets is$0$precisely if the sets are equal. For$A\Delta B=\emptyset$if and only if$A=B$. 1 Assume$C$is compact. Take a covering of$C$by balls$B_\rho(x)$such that$x$is in the complement of$D$and also the ball$B_\rho(x)$is in the complement of$D$. This is possible as the complement of$D$is open. Now you can select a finite number of such balls covering$C$. These balls do not touch$D$. Thus the distance between$C$and$D$is ... 1 Suppose that$d(C,D)=0$and that$C$is compact. So there exists a sequence of elements$c_n\in C$such that$d(c_n,D)\to 0$. Because$C$is compact,$\{c_n\}$admits a convergent subsequence, say$\{c_{n_k}\}$, that converges to$c\in C$. But then $$d(c,D)=\lim d(c_n,D)=0.$$ That is, there exists a sequence$\{d_n\}$in$D$with$d_n\to c$. As$D$is ... 1 Let$\bar f:\bar X_1 \to \bar X_2$constructed with limits of Cauchy sequences. It remains to show that$\bar f$is an isometry. Let$a,b\in \bar X_1$be given. Then there there are sequences$(a_k), (b_k)$in$X_1$converging to$a$and$b$, respectively. Moreover, it holds $$\bar f(a) = \lim_{k\to\infty} f(a_k), \quad \bar f(b) = \lim_{k\to\infty} ... 1 Generally if you have a sequence x_n \to x and x_n \in F a closed set for all n, then x \in F. In this case, take F=[0,\infty), then if we let x_n=d(a_n,b_n)+d(b_n,c_n) - d(a_n,c_n) we see that x_n \in F for all n and hence x = B+C-A \in F, or, in other words, x \ge 0 and so A \le B+C. 1 Since you didn't state which definition of openness you're referring to, i'll use this one: A set A is open if for all a \in A there exists an open (bounded) ball B such that a \in B \subseteq A. Now, let B' = \{ y : d(x,y) > r\} and b \in B'. Let r' := d(x,b) -r > 0. Then we claim that \{y: d(b,y) < r'\} \subseteq B'. Take y such ... 1 HINTS:$$\between\bullet\quad\bullet$$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$S=\{x:f(x)=g(x)\}$.to show that$S^c$is open . Let$a\in S^c\implies f(a)\neq g(a)$.Let$d(f(a),g(a))=r$. Since$Y$is Hausdorff so$\exists r>0$such that$B(f(a),\frac{r}{4})\cap B(g(a),\frac{r}{4})=\emptyset$. Since$f,g$are continuous then$\exists \delta_1,\delta _2>0$such that$f(B(a,\delta _1))\subseteq B(f(a),\frac{r}{4})$and ... 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. 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 I'm assuming that$d(f,g)=\sup\{f(x)-g(x):x\in X\}$. If so, convergence with respect to this metric is uniform convergence. 1 Well, when you say by Supreme metric space, it does mean uninformative convergence. Of course uninformative convergence is stronger than point-wise convergence.The norm of$B(X)$is$ \Vert f\Vert_{\infty} = \sup\limits_{x\in X} \vert f(x)\vert. \$ So by your assumption, it is uninformative. For more information, please refer to the "Principles of real ...

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