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3

$X=[0,1]$ with the usual metric, $\phi(x)=\sqrt x$.


3

The idea is OK, but should be formalised a bit more. Let $Y = [1,\infty)$ with the Euclidean metric $d$. Clearly, $Y$ is complete as it is a closed subset of the complete $\mathbb{R}$ in the Euclidean metric. Then $f(x) = \frac{1}{x}$ from $(X,e)$ to $(Y,d)$. Then $d(f(x), f(y)) = |f(x) - f(y)| = |\frac{1}{x} - \frac{1}{y}| = e(x,y)$ for all $,y \in X$. ...


2

Suppose not, we have a subsequence of $f(x_n)$ s.t. $d'(f(x_{n_k}),f(x))\ge \epsilon$.WOLG, we assume the subsequence is $f(x_n)$ and $x_n\not =x\forall n$ (otherwise restrict our discussion to suitable subsequence). Then set $B=\{f(x_n)\}$, hence $ \{(x_n)\}\subset f^{-1} ( B),\{(x_n)\}\cup \{x\}\subset \overline{f^{-1} ( B)}$ since $x_n\to x$ By ...


2

For $x,y\in\mathbb{R}^n$, the set $\{\lambda x+(1-\lambda)y\mid 0<\lambda<1\}$ is the open line segment from $x$ to $y$. Thus your definition says that if $x,y\in E$ (convex), then the entire line segment joining $x$ to $y$ is also in $E$.


2

SKETCH: Suppose that you have metric spaces $\langle X_i,d_i\rangle$ for $i\in I$. First replace $d_i$ be $d_i'$, where $$d_i'(x,y)=\min\{d_i(x,y),1\}$$ for $x,y\in X_i$. Observe that $d_i'$ generates the same topology as $d_i$ and is complete if $d_i$ is complete. Now define a metric $d$ on $\bigsqcup_{i\in I}X_i$ by $$d(x,y)=\begin{cases} ...


2

No. Let $X=(0,1], d_1=|x-y|,d_2=|\frac{1}{x}-\frac{1}{y}|$, you can check they are topologically equivalent. But we don't have strong equivalence as $x,y\to 0$


2

The closure of $E$, in a topological space $X$ containing it, is $E$ together its limit point. This proves your statement: "E is closed if every limit point of E is a point of E" which is true. Conversely, a closed subset of a topological space can have no limit points: for example $\{0\}$ is closed in $\mathbb R$, but it hasn't limit points; it has only ...


2

One way of proving that norms give the same topology is to find inequalities $$c||x||_1\leq ||x||_2\leq C||x||_1\ \ \ \ \text{ with }c>0.$$ $$\text{If }\ \ q>p\geq1\ \ \ \ \ \ \ \ \ \text{ then }\ \ \ \ \ \ \ \ \ \frac{n^{1/q}}{n^{1/p}}||x||_p\leq||x||_q\leq||x||_p.$$ The first inequality is the generalized mean inequality and the second is because ...


2

I will assume that you meant $f(x) = g(a)$ as a hypothesis in the second line, instead of $f(x) = f(a)$, otherwise I don't think it will work. Comments: Since $f,g$ are continuous $x∈B(a,δ)⟹f(x)∈B(f(a),ϵ)$ and $x∈B(a,δ)⟹g(x)∈B(g(a),ϵ)$ (in this case I'm already considering minimum delta that satisfies this). This is correct. However, I do recomend ...


2

Your argument is incorrect. There are many more non open subsets of $(a,b)$, besides sets of isolated points. You're basically assuming the thesis to prove it, because intervals are indeed the only connected subsets of $(a,b)$ (or, more generally, of $\mathbb{R}$).


2

If $d_V(x,y) = \dfrac{1}{m!}$ and $d_V(y,z) = \dfrac{1}{n!}$, then $m!$ divides $|x-y|$ and $n!$ divides $|y-z|$. In particular $m!$ divides $x-y$ and $n!$ divides $y-z$. Assume without loss of generality that $m \le n$. Then $m!$ divides $n!$, so that $m!$ divides both $x-y$ and $y-z$. Thus $m!$ also divides $x-z$. It follows that $m!$ divides $|x-z|$ ...


1

Think that any linear combination of $x$ and $F$ will be closed for any $x \in \Bbb R^n$!! i.e. $ \alpha x + \beta F $ will be closed for scalars $\alpha , \beta$.


1

$$\mathbb{R}-\mathbb{Z} = \cdots \cup (-3,-2) \cup (-2,-1) \cup (-1,0) \cup (0,1) \cup (1,2) \cup (2,3) \cup \cdots$$ This is a union of open intervals, and is therefore open. So its complement $\mathbb{Z}$ is closed.


1

The set of all polynomials with rational coefficients is a countable dense subset of the set of all polynomials with real coefficients. Hence it is also a dense subset of $C^1$ by your argument. To see that it is countable, you can check that there is a bijection with $\bigcup_{n=1}^\infty Q^n $ which is a countable union of countable sets.


1

A (mathematical) space is a set endowed with certain "structure" on it. It doesn't come quite often to talk about the distinctions between "space" and "structure" in daily studies since they refer to different aspects of points of interest. For example, a vector space is a set endowed with the algebraical structure, and a topological space is a set endowed ...


1

Your approach works if you combine your two ideas: Form an open set around each point in $K$, such that it contains only finitely many points of $S$. These open sets cover $K$, and since $K$ is compact, extract a finite subcover, each element of which only contains finitely many points of $S$.


1

Let's call $B=A\cup \{0\}$ and we define $d:B\times B\to B$ as $d(a,b)=0$ if $a=b$ and $d(a,b)=\max\{a,b\}$ if $a\neq b$. Note that $d(0,b)=b$ for each $b\in B$ so $d$ is indeed a surjection. The other axioms for a metric are easily verified (there are several cases though).


1

The notation isn't the best. I would write the theorem like this: if $\phi$ is a linear functional on $\mathcal{F}$ such that, for any unit vector $u\in\mathcal{F}$, $\frac{\phi[tu]}{|t|}\to0$ as $|t|\to0$, then $\phi=0$ identically. But written like this it becomes obvious since $\frac{\phi[tu]}{t}=\phi[u]$, so the condition pretty much says that ...


1

You are given a space ${\cal F}$ and a linear functional $\phi:\>{\cal F}\to{\mathbb C}$ that you are trying to better understand. The only thing you are told is that $$\lim_{h\to0}{\phi(h)\over\|h\|}=0\ .$$ The statement in question says that in such a case one necessarily has $\phi(x)\equiv0$ on ${\cal F}$. For a proof consider an arbitrary ...


1

Using the definition of openness that was given in your previous question. Note that the complement of $[a,b]$ is $]\infty,a[ \cup ]b,\infty[$. For any point in this set, you can find a small open ball that is contained within the set.


1

Hint The metric you are looking at is the standard euclidean metric. A set $A$ is open in a metric space $(X,d)$ iff for any $a\in A$ there exists an $\epsilon > 0$ such that the set $$B_\epsilon(a) := \{x\in X | d(a,x) < \epsilon\}$$ is contained in $A$. Now $(0,1)$ is open in $(\mathbb R, |\cdot |)$, so for any $t\in (0,1)$ there exists a $\delta ...


1

For $0<p<1$ the $p$-norm is only a quasinorm. Nevertheless, it is not hard to check that all of them give the same topology as the $\infty$-norm $$||(x_1, \ldots, x_n)||_{\infty} = \max|x_i|$$ Let $x$ with $||x||_{\infty} < \epsilon$. Then $\sum_{i=1}^n |x_i|^p \le n \epsilon^p$ and so $||x||_{p} < n^{\frac{1}{p}}\cdot \epsilon$. Therefore ...


1

The definition is correct as is: "$E$ is closed if every limit point of $E$ is a point of $E$." To see how a limit point could not be a point of $E$, consider the (open) interval $E=(0,1)$. Then for any $\epsilon$ neighborhood of zero, there is some positive number ($\min\left(\frac{\epsilon}{2},\frac{1}{2}\right)$ for instance) which is contained in the ...


1

Hints: Are you familiar with the discrete metric? What is the distance of the points $x=(0,0)$ and $y=(0,1)$? The only thing to check is the triangle inequality. Note that $\delta(x,y)=|x_1-y_1|+|x_2-y_2|$ defines a metric. Now observe that the function $\phi(x)=\sqrt x$ is increasing and satisfies $\phi(x+y)\leq\phi(x)+\phi(y)$ for all $x,y\geq0$. Can you ...


1

A subset G of Y is relatively open in Y if and only if there is an open subset U in X with G = U ∩ Y


1

1) if $y\in f(A_0)$ then $y=f(x)$ for some $x\in A_0\subset A_1$, an thus $y\in f(A_1)$. 2) Let $y\in f(A_0\cup A_1)$, then $y=f(x)$ for a $x\in A_0\cup A_1$ and thus $y=f(x)$ for a $x\in A_0$ or for a $x\in A_1$. The reciprocally is the same. 3) Let $y\in f(A_0\cap A_1)$, then $y=f(x)$ for a $x\in A_0\cap A_1$ and thus $y=f(x)$ for a $x\in A_0$ and for ...


1

It seems the following. Let $\{X_\alpha\}$ be a family of (completely) metrizable spaces. Then for each $\alpha$ there exists a (complete on $X_\alpha$) compatible metric $d_\alpha$. Then $$d(x,y)=\cases{ \min\{d_\alpha(x,y),1\},\text{ if }x,y\in X_\alpha\mbox{ for some }\alpha\\ 2,\mbox{ otherwise }} $$ is a (complete) compatible metric on the direct ...


1

Recall the following theorem. Theorem. Let $X$ be a subspace of a topological space $Y$ and let $E\subset X$. Then $E$ is closed in $X$ if there exists a set $W$ closed in $Y$ such that $E=X\cap W$. The proof of this theorem is not difficult and worth writing down yourself and good practice for thinking about subspaces. If you accept this theorem, your ...


1

You need to fill in some important details, but the basic idea is fine. You might start like this: Let $\langle x_n:n\in\Bbb N\rangle$ be a Cauchy sequence in $\langle X,e\rangle$. For $n\in\Bbb N$ let $t_n=\frac1{x_n}$. Then by definition for each $\epsilon>0$ there is an $n_\epsilon\in\Bbb N$ such that $|t_m-t_n|<\epsilon$ whenever $n\ge ...



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