New answers tagged

0

Here is an example with one more property: $X$ is complete. Namely, let $X$ be the real line with the metric $d(x,y)=\min(|x-y|,1)$. At small scales this is the same as Euclidean metric, so the space is complete and locally compact. It is also $\sigma$-compact, as the union of intervals $[n,n+1]$. On the other hand, it is not proper since any closed ball ...


1

If $f[X]$ is proper, pick $x_0 \in X, x_0 \notin f[X]$. Then $d(x_0, f[X]) = r > 0$, as $f[X]$ is also compact, thus closed. Then define $x_{n+1}= f(x_n)$ for $n\ge 0$, then (as all other $x_n$ besides $x_0$ are $f$-images) we have $d(x_0, x_n) \ge d(x_0, f[X]) \ge r$ for all $n \ge 0$. Hence, as $f$ is an isometry, $d(x_k, x_{n+k}) = d(x_0, x_n) \ge ...


1

I think expecting the same intuitive result for open balls in different metric spaces isn't a good idea in general. For example, take $\Bbb R$ with the discrete metric $d(x,y)=0$ iff $x=y$ and $d(x,y)=1$ otherwise. Then $B(0,r)=\Bbb R$ if $r>1$ and $\{0\}$ if $r\leq 1$. Of course it's quite unintuitive that what's in the ball doesn't depend on the ...


0

Just work with the definition of uniform continuity (you need it, as BigbearZzz's answer shows). I write $|x-y|$ and $|f(x)-f(y)|$ to simplify the notation. Assume $f$ is uniformly continuous, and let $(x_n)$, $(y_n)$ be two sequences with $|x_n-y_n|\to0$ $(n\to\infty)$. Let an $\epsilon>0$ be given. Then there is a $\delta>0$ with ...


3

Yes, in any metric space, you may define the ball of any center and radius. In your example, the ball of center $0$ and radius $1$ in the metric space $[0,1]$ is $[0,1)$.


4

I'd say so yes. The ball just denotes the set $B(0;1):=\{x\in X\mid d(0,x)\leq 1\}$. It has nothing to do with the geometric intuition one might have about balls


2

Let $X=\{\frac{1}{n}:n\in\mathbb{N}\}\bigcup\{0\}$. Consider two metric on $X$: $d_{1}=$usual Euclidean distance, which gives subspace topology viewed as a subset of real line. The components of $(X,d_{1})$ are singletons. (Why?) $d_{2}=$discrete distance, which gives discrete topology, every subset is an open set. The components of $(X,d_{2})$ are also ...


2

Unfortunately no, you are not on the right track. Suppose $X:=\Bbb R$, let $x_n:=n$ and $y_n:=n+\frac 1n$, then $$ d(x_n,y_n)=\frac 1n $$ which converges to $0$ as $n\to \infty$. However, there is no such $x$ that $x_n\to x$. You need to use the full power of uniform continuity to prove the statement, continuity is not enough.


1

For the first proof, if you have a separation, $M = A \cup B$, then $A$ and $B$ are both open, and $A \cap B = \emptyset$. But, $A$ is also closed since $B$ is open, and $A = M \setminus B$. Same goes for $B$. So this is the contrapositive of the reverse direction of the statement. When they say reciprocally, they are referring to the forward implication ...


0

If $M=A \cup B$ is a separation, both $A$ and $B$ are open as you say. Then $A = B^c$ is the complement of an open set, so is closed. The recipricolly works the same way. If $A$ is open and closed, $M \setminus A=B$ must also be open and closed as the complement of a clopen set.


0

Assume the first characterisation is true. Then we want to show $f(B(a, \delta))\subseteq B(f(a), \epsilon)$, so take any element $f(x)\in f(B(a, \delta))$. As $x\in B(a, \delta)$, it must be less than $\delta$ away from $a$, and so the first characterisation says that the distance between $f(x)$ and $f(a)$ must be less than $\epsilon$. So $f(x)$ is ...


0

First of all, a correction: $f$ is continuous if for every $x$ and for every $\epsilon$, there exists a $\delta$ such that if $d(x,a)<\delta$, then $d(f(x),f(a))<\epsilon$. Hints: One set, $A$, is a subset of another, $B$, if it is true that for every $x\in A$, $x$ is also an element of $B$. $x$ is an element of $B(a, \delta)$ if and only if $d(x,a) ...


1

$B(a,\delta)=\{x\colon d(x,a)<\delta\}$ $B(f(a),\epsilon)=\{y\colon d(y,f(a))<\epsilon\}$ the continuity condition: $d(x,a)<\delta \implies d(f(x),f(a))<\epsilon$ means that for every $x$ such that $d(x,a)<\delta$ you have $d(f(x),f(a))<\epsilon$, what is $f(x)\in \{y\colon d(y,f(a))<\epsilon\}=B(f(a),\epsilon)$


3

The result is not true. Let $M=\mathbb R$ and $f(x)=\max(x,0)$. Then $A=(0,\infty)$ and $\partial A=\{0\}$, while $\{x:f(x)=0\}=(-\infty,0]$. What is true is $\partial A\subset\{x:f(x)=0\}$.


0

For each subset $I$ of the positive integers $\mathbb{N}$, define $e_I\in\ell^\infty$ by $$ (e_I)_i=\begin{cases} 1,&i\in I\\ 0,&1\not\in I \end{cases} $$ Then $d_\infty(e_I,e_J)=1$ whenever $I\neq J$ (why?). So $$ \mathbb{B}=\{B(e_I,\frac12):I\subset\mathbb{N}\} $$ is an uncontably infinite (why?) collection of disjoint (why?) open balls in ...


3

No. Consider that this problem is equivalent to finding the roots of $f$ (if you knew the distance from $x$ to the nearest root, you could check only two points to find the root). If you know the roots and they are countable, you can of course simply sort them and quickly find the one closest to $x$. As for finding the distance in practice, you can ...


3

Show two inclusions, based on your definition of closure as the union of interior and boundary: If $x \in \operatorname{Int}(A)$ then in particular $x \in A$ and so $d(x,A)$ is the infimum of a set that contains $d(x,x) = 0$ in particular, so $d(x,A) = 0$. If $x \in \partial A$, pick $r>0$ arbitrarily. Then $B(x,r)$ intersects $A$ by the definition of ...


1

Let $x\in \bar A$ s.t. $d(x,A)=m>0$. By definition of $\bar A$, there is a sequence $(x_n)\subset \bar A$ s.t. $x_n\to x$ when $n\to \infty $. Let $\varepsilon<m$. Then, there is a $N$ s.t. $d(x_N,x)<m $, which is a contradiction with the fact that $d(A,x)=m$. Therefore, $d(x,A)=0$.


0

It's OK, but there are some notation remarks: Start by saying "let $\left\{f_\lambda: M \rightarrow \mathbb{R} \,|\, \lambda \in L \right\}$ be a family of continuous functions (as @user254665 already said)". say "then $F_\lambda := f_\lambda^{-1}\left[[0,+\infty)\right]$ is closed" (not $f$). The next line is then unnecessary (we defined $F_\lambda$ in ...


0

Of course not necessarily, since in general $G_\delta$-sets are not preserved by continuous (and even open) maps (and even retractions). Let $X=\mathbb R^2\times\mathbb N$ be the product of spaces $\mathbb R^2$ and $\mathbb N$ endowed with standard topologices. Put $Y=\mathbb R^2\times\{1\}\subset X$ and $r:X\to Y$ be the projection onto the first ...


1

By contradiction, suppose $f$ is discontinuous at $x.$ Then, from the $\delta$ -$\epsilon$ def'n of continuity, $$\exists \epsilon >0 \;\forall \delta >0\;\exists y\;(|x-y|<\delta \land |f(x)-f(y)|\geq \epsilon).$$ So take such an $\epsilon,$ and for each $n\in N,$ take $y_n$ such that $|x-y_n|<1/n$ and $|f(x)-f(y_n)|\geq \epsilon.$ Now for ...


2

Let $(x_n)$ be a convergent sequence in $M$ with limit $a$. Then we can consider the sequence $x_1, a, x_2, a, x_3, a, \ldots$ which intertwines $(x_n)$ with the constant sequence $(a)$. This sequence converges to $a$ as well, so by assumption, the sequence $f(x_1), f(a), f(x_2), f(a), \ldots$ converges in $N$. Since every other term is $f(a)$, the only ...


0

First: If $N$ is a metric space, then the diagonal $\Delta_N = \{(y,y) \mid y \in N\}$ is a closed subset of the Cartesian product $N \times N$. (HINT: If $(y,z) \not \in \Delta_N$ then $d(y,z) =: d > 0$; then the Cartesian product of balls of radii $d/2$ centered at $y$ and $z$ respectively is an open set in the Cartesian product of $N$ with itself, and ...


0

For the final example, consider $M=[0,2), N=\mathbb{R}, A=[0,1), B=[1,2)$. Then, take $f(x)=x$ when $x \in A$, and $f(x)=x+1$ when $x \in B$. Obviously $f$ is continuous when restricted only to $A$ or $B$, but there's a discontinuity at $x=1$ when examining it over all of $M$.


2

Consider $X = \mathbb{Z}$ with a metric $d$ such that $d(m, n) = 1$, if $m \neq n$ and $d(n, n) = 0$. $d$ is a metric. Take any infinite subset $A$ (for example, $\mathbb{N}$). Then $A$ is open (and closed) and it is also bounded (because the maximum distance is 1). Moreover, $A$ is not compact, because the family of sets defined by $\{\{x\} \mid x \in ...


1

HINT: Start with any non-compact metric space, and define a new metric $d'$ by setting $d'(x,y)=\min\{d(x,y),1\}$ for $x,y\in X$; $d'$ generates the same topology as $d$.


6

$X=(0,1)$ with Euclidean metric. Then $X$ itself is closed and bounded yet not compact.


0

Warning: this proof is full of the most disgusting indexing the world has ever known. Wait, actually in retrospect this proof has less disgusting indexing than the proof for $L_\infty$. Suppose we have a sequence of points $x_n$. For each $x_n$, we will denote the $t^{th}$ element of the sequence-vector (recall that each point is a sequence-vector) as ...


1

You're almost there. Pick any $r > 0$. We also have $r' > 0$ such that $C(x,r') \cap B = \emptyset$. So for $r'' = \min(r,r') > 0$ we have that $C(x,r'') \cap A \neq \emptyset$, and any point in this intersection cannot be in $B$, by the property of $r'$. And this point is certainly in $C(x,r)$ as well, so this ball intersects $A \setminus B$. ...


0

Assume (vii), and let $i:X\to X$ be the identity function. Clearly $i$ is continuous as a function from $\langle X,d\rangle$ to $\langle X,d\rangle$, so by the hypothesis (vii) it is continuous as a function from $\langle X,d\rangle$ to $\langle X,e\rangle$, and (iv) holds. Now assume (vi). The function $i$ is continuous as a function from $\langle ...


0

(vi) Every function from $X$ into a metric space that is continuous with respect to $e$ ... Take the function $\text{id}:(X,e)\to(X,e)$, $(X,e)$ being the arbitrary metric space. Then, by (vi), $\text{id}:(X,d)\to(X,e)$ is continuous. Similarly with (vii), take $(X,d)$ as the arbitrary metric space. My idea for this answer was thanks to Tim Huijgens ...


3

Any metric space is a topological space. So topological terms generally have the same meaning as in a general topological space. In particular a metric space is separable if it has a countable dense set.


1

Yes, it is true. Suppose it was not true, then for every integer $n>0$, you have $x_n\in K_{1/n}$, $x_n$ is not in $G$. There exists $y_n\in K$ such that $d(x_n,y_n)\leq 1/n$. Since $K$ is compact, you can suppose (up to a subsequence) that $(y_n)$ converges towards $y$, $d(y,x_n)\leq d(x_n,y_n)+d(y_n,x_n)$ implies that $x_n$ converges towards $y$. This ...


1

Hint: Since $G$ is open, for every $x\in K$ there exists $\epsilon_x$ with $B(x,\epsilon_x)\subset G$. Now $$K\subset\bigcup_{x\in K}B(x,\epsilon_x/2).$$


1

The following is not so much an answer to the question, as a commentary on the unnecessary use of the axiom of choice in the proof given by the author of the question. I am doing this because this kind of abuse of AC is typical. We can avoid using AC in the following way. Suppose we have a compact subset $A$ of a Hausdorff space $X$ (every metric space has ...


3

Your original question "Construct a metric such that the resulting metric topology is the same as the standard topology" makes no sense since, as Donkey_2009 was trying to tell you, you have to have a metric to begin with in order to define "radius" and "ball". I suspect your problem was to construct another metric which gives the same topology. That's ...


1

Your proof is correct - assuming you know how to show that $G_1$ and $G_2$ are indeed disjoint. You only used that compact subsets of metric spaces are closed in your proof, so you have effectively proved the stronger assertion that metric spaces are completely regular/$T_{3\frac{1}{2}}$. Using the hint, you would prove a different strengthening of the ...


2

You're a little muddled about what open and closed mean. In a metric space, open and closed sets are defined using the concept of a ball. This proof does not deal with metric spaces, but with topological spaces, which are more general, so there is no such thing as a ball here. Here, to know whether the two halves of the separation are open, you just need ...


1

Let $f: A \to N$ and $b:= \lim_{x \to a} f(x)$. If $b \in f(A)$ you are done, so we will suppose not. Let $N_b$ be any neighborhood of $b$. Suppose $N_b \cap f(A) = \emptyset$. By definition of $b$, given any neighborhood $U_b \subset N$, there exists a corresponding $V_a \subset A$ s.t $f(V_a) \subset U_b$. Now just take $N_b = U_b$ and you have a ...


0

HINT: Prove the contrapositive instead: suppose that $A$ is not open, and show that there are some $x\in A$ and some sequence $\langle x_n:n\in\Bbb N$ contained entirely in $M\setminus A$ that nevertheless converges to $x$. This is fairly straightforward and logically equivalent to what you want.


2

So this doesn't quite work but you are definitely on the right track. The only problem with your argument is that we don't know a priori that $\frac{1}{\epsilon}$ is a natural number. However, try repeating your argument with any natural number $k$ such that $k<\frac{1}{\epsilon}$. This will solve the problem, because given $\epsilon>0$, we have ...


3

Since $n_0\in\mathbb N$, you can't choose $n_0=\frac{1}{\varepsilon}$. What you can choose is $n_0=\lfloor\frac{1}{\varepsilon}\rfloor +1$. But you don't need to explicit the $n_0$. Let $\varepsilon>0$. Since $$\lim_{n\to \infty }\frac{1}{n}=0,$$ there is a $n_0\in\mathbb N$ s.t. $\frac{1}{n}<\varepsilon$ when $n\geq n_0$. Therefore, if $n\geq n_0$, ...


1

HINT: Show that $d$ is equivalent to the usual metric on $\Bbb R$, meaning that it generates the same topology on $\Bbb R$. Compactness is a topological property, not a metric property, so as long as two metrics generate the same topology, a set compact in one is compact in the other. Completeness then comes for free. It’s also possible to show directly ...


3

Real numbers with the discrete metric. I.e. $d(x,y) = 1$ if $x \neq y$.


3

You need to do this for the standard metric on the reals. If not, it should have been specified, as Cauchy-ness depends on the metric used, and it can fail for other choices of the metric on the reals. So you need to show that for every $\varepsilon > 0$ there exists some $N$ such that for all $n,m \ge N$ we have $$\left| d(x_n, y_n) - d(x_m,y_m) ...


0

The intended metric in $\mathbb{R}$ is probably the usual metric here, namely the one defined by the absolute value of the difference. The proof is particularly simple if you are allowed to use a metric completion $\mathbf{X}$ of $X$. Passing to the limits $x=\lim_{n\to\infty}x_n$ and $y=\lim_{n\to\infty}y_n$ you can easily show from the triangle ...


0

It is a correct approach, except that it should be $\frac{2}{l+1}$. But note that your proof as stated uses the axiom of dependent choice, because you can't uniquely choose the subsequence in the ball of desired radius if there are more than one and you don't have any way of tie-breaking.


2

Let $\mathscr{C}$ be the collection of $E\in\mathcal{B}_X^*\otimes \mathcal{B}_Y$ for which: There exists $A\subseteq X$ null with $E\cup (A\times Y)\in\mathcal{B}_X\otimes\mathcal{B}_Y$. There exists $A\subseteq X$ null with $E\setminus (A\times Y)\in\mathcal{B}_X\otimes\mathcal{B}_Y$. Then it suffices to show that $\mathscr{C}$ is a $\sigma$-algebra ...


3

Let $n \in \mathbb{N}$, and define $ f_n(x) = \left\{\def\arraystretch{1.2}% \begin{array}{@{}c@{\quad}l@{}} 0 & \text{if } 0 \leq x \leq 1/2\\ 2nx - n & \text{if } 1/2 < x \leq (n + 1)/(2n)\\ 1 & \text{if } (n + 1)/(2n) < x \leq 1\\ \end{array}\right.$ Then $\{f_n\}$ is a Cauchy sequence whose limit is $ f(x) = ...


1

There are at least two reasonable approaches. One is to construct a complete metric equivalent to the usual one; I’ll get to that in a moment. The other is to prove the result that a $G_\delta$ set in a complete metric space is completely metrizable; since $[0,1)$ is certainly a $G_\delta$ in $\Bbb R$, that theorem yields the desired result. To get a ...



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