New answers tagged

1

The question seems a bit unclear to me, but I guess you are asking: How to see that the Baire space has a countable base. The standard base for the product topology on $\newcommand{\N}{\mathbb N}\N^{\N}$ is the base consisting of sets $$\mathscr N_{a_0,a_1,\dots,a_n}=\{x\in\N^{\N}; x_0=a_0, x_1=a_1,\dots,x_n=a_n\}.$$ I.e. the basic sets consist of sequences ...


0

If that were true, then any countable T$_1$-space would be metrizable. In fact there are lots of countable Hausdorff spaces that aren't even first countable. For example, topologize $\mathbb N$ so that $$S\text{ is closed }\iff1\in S\text{ or }\sum_{n\in S}\frac1n\lt\infty.$$


3

Your notation suggested another alternative: Let $g^2 = (g^1)^2$ -- a local extremum of $g^1$ is also a local extremum of its square. In fact, let $\phi$ be strictly increasing on $[ 0, \infty )$, then $g^2 = \phi(g^1)$ replicates the extrema of $g^1$. So geodesics (paths whose length is first order stationary under perturbation, hence correspond to local ...


4

No. Consider as your model set $M$ a tripod. (A tripod is a graph with one vertex of degree three and three vertices of degree one attached to it.) For your different length metrics, just assign the edges of the tripods various different lengths, and equip the space with the length metric induced by those edge lengths (making each edge into an isometric ...


1

You know that given $\epsilon>0$ there is $N$ such that $$ |f_n(x)-f_{n+p}(x)|\le\epsilon\quad\forall x\in E,\quad\forall p\in\mathbb{N}.\tag{1} $$ You also know $\lim_{n\to\infty}f_n(x)=f(x)$. Now let $p\to\infty$ in (1).


1

All correct (I think I wrote that on wikipedia) $$d(O, Q) = \frac{1}{2} \ln \frac{1+r}{1-r} = \operatorname{artanh} r$$ see also the formulas for tanh and artanh at en.wikipedia.org/wiki/Hyperbolic_function I have added it to the wikipedia page ps related: For the Poincare disk model he hyperbolic distance between O the center of unit disk and a point ...


3

It is clear that $d$ is defined on $\mathbb R^2$, because no matter what $u_1,u_2, v_1, v_2$ are, you can calculate $d(u,v)$. You show that $d$ is a metric by showing that it satisfies all the properties that metrices must satisfy: Show that $d(u, v)\geq 0$ for all $u,v$. Show that $d(u,v)=0 \implies u=v$. Show the triangle inequality.


0

For each $t\in K$ let $M_t$ be open in $M$ and let $K_t$ be open in $K $ such that $$ (a,t)\in M_t\times K_t\subset V.$$ Since $\{K_t:t\in K\}$ is an open cover of the compact space $K,$ we can take a finite $F\subset K$ such that $\cup \{M_t :t\in F\}= K .$ Now let $U=\cap_{t\in F}M_t.$ Then $a\in U\subset M_t$ and $M_t\times K_t\subset V $ for each $t\...


2

Assuming that you’ve reproduced it accurately, that proof makes life difficult by using $d$ both for the metric on $M\times K$ and for the metric on $M$. In fact the metrics just get in the way: the result is true for an arbitrary space $M$ and an arbitrary compact space $K$, and I think that the proof is actually a bit clearer in that setting, so let me ...


2

It’s basically correct. There’s a typo at the very beginning, where you meant to write ‘For each $n$ there exists’ (instead of $j$). And you need to pass to a convergent subsequence only once: the tail sequence $\langle x_n:n\ge k(j)\rangle$ already converges to $\hat x$, since it’s a subsequence of a sequence converging to $\hat x$. For a proof in case $X$ ...


1

Your reasoning for $d(A,B)=1$ is flawed. $B$ does not need to be a subset of the $2\delta$-interval centered at $0$. Instead, $B$ needs to be a subset of $A_\delta$ which is equal to the union of $2\delta$ intervals centered on all the points of $A$, that is $$B \subset [0-\delta,0+\delta] \cup \bigcup_{n \in \mathbb{N}} \,\, \biggl[\frac{1}{n}-\delta,\...


4

No. Let $X=[1,\infty)$. Define $$d_1(x,y)=|x-y|,\quad d_2(x,y)=\left|\frac 1x-\frac 1y\right|.$$Then $d_1$ is certainly complete, but $(x_n)$ is a Cauchy sequence with respect to $d_2$ if $x_n=n$.


1

Let $(X,d_X)$ and $(Y,d_Y)$ be two metric spaces, with $X$ compact, and let $f:X\to Y$ be a continuous bijection. Definition: The function $f$ is continuous if for every set $O \subset Y$ which is open in $Y$, $f^{-1}(O)$ is open in $X$. Lemma: The function $f$ is continuous if and only if for every set $C \subset Y$ which is closed in $Y$, $f^{-1}(C)$ is ...


3

$T$ cannot be injective. If $T$ is constant this is obvious, so assume $T$ is not constant. Then, as you explained already, $T$ attains its minimum an maximum in $x_0$ and $x_1$, say, respectively. Now choose two different paths from $x_0$ to $x_1$ which are disjoint with the exception of the common endpoints. The image of each covers the interval $[T(x_0), ...


0

You're confused about the definitions here, and what is given vs. what is to be proved. You are assuming that $M \times N$ is compact. This means (by definition) that given any open cover $\mathscr{U} = \{U_\lambda: \lambda \in \Lambda\}$ of $M \times N$, we can find a finite subover of $\mathscr{U}$. From this we want to show that $M$ is compact and $N$ ...


0

The set of all points S = {$\phi$(x)|$x\in M$}= ($-\infty$,$\epsilon$] is clearly closed in $\mathbb R$ because $S^c$ = ($\epsilon$,+$\infty$),which is open in the usual topology on $\mathbb R$. $\phi$ is continuous by definition. Therefore $\phi^{-1}(S) \subseteq M$ is closed because the image is contained in the closed interval S $\subseteq\mathbb R$ and ...


2

The inverse image of a closed set under a continuous map is closed and since $\{y \in \mathbf{R}: y \leq \epsilon\}$ is closed, it follows that $\{x \in M: \phi(x) \leq \epsilon\}$ is closed.


0

Denote by $d(x,y):=\sup_{0\leq t\leq T}|x(t)-y(t)|$ the "usual" distance in $X:=C([0,T])$. Then $$\rho(x,y)\leq d(x,y),\quad d(x,y)\leq e^{LT}\rho(x,y)\qquad\forall x,\>y\in X\ .$$ This shows the Cauchy sequences in $(X,\rho)$ and $(X,d)$ are the same. Since $(X,d)$ is known to be complete (see below) we can at once infer that $(X,\rho)$ is complete as ...


6

So, I must prove that if I have an open cover in $M\times N$, with a finite open subcover, I must have an open cover in $M$ and $N$ with finite open subcovers. However, I don't have any idea on how to prove that. No, that is not what you must prove. What you must prove is that assuming that $M\times N$ is compact, if you have an open cover of $M$, then it ...


1

I won't insert all the details into a hint, but consider this oft-used technique: $|x(t_2)-x(t_1)|<|x(t_2)-x_m(t_2)|+|x_m(t_2)-x_m(t_1)|+|x_m(t_1)-x(t_1)|$, the RHS of which is arbitrarily small. The existence of the function $x(t)$, the 'limit' of the Cauchy sequence, or more formally that function for which $\sup_{0\leq t\leq T}|x(t)-x_m(t)|$ is ...


0

For all $x\in X$, we have $\rho(f_n(x),f(x))\geqslant \rho(f_{n+1}(x),f(x))$. It follows that $$\sup_{x\in X}\rho(f_n(x),f(x))\geqslant\sup_{x\in X} \rho(f_{n+1}(x),f(x)).$$ Since $\lim_{n\to\infty} \rho(f_n(x),f(x))=0$ for all $x\in X$ we conclude that $$\lim_{n\to\infty}\sup_{x\in X} \rho(f_n(x),f(x))=0,$$ and hence $f_n$ converges uniformly.


0

Not sure about what you meant by saying "achieves a local maximum". If you meant that $M$ is the set of local maximum points, then here is a counter-example. Let us consider the following function $f:R\rightarrow R$. $f(x) = 0$ for $x \leq 0$ and $x\geq 1$, $f(x) = x$ for $x > 0$ and $x \leq 1$, $f(x) = 1$ for $x > 1$ and $x \leq 2$, $f(x) = x-1$ ...


1

Writing $(-\infty,-1]\cup\{-1\}$ is redundant since $-1\in(-\infty,-1]$. When you write cl in LaTeX, write \operatorname{cl} instead; it will give it the right styling. Lastly, $\operatorname{cl}\mathbb{Q}=\mathbb{R}$, since each sequence in $\mathbb{Q}$ converges to a real number and any real number can be written as a convergent sequence of rationals.


1

I would like to say a few things here...are you studying analysis or topology? If you are studying analysis then you can talk about closure which directly means closure in standard topology but if you are studying topology you need to ask first is it standard topology or discrete topology or lower limit topology or K topology and stuff like that. Assuming ...


5

No, it's possible to construct an uncountable closed set containing only irrationals. Let $\{p_n\}$ be an enumeration of $\mathbb{Q}$. Put $I_n = (p_n - 2^{-n-1},p_n+2^{-n-1})$. Then each $I_n$ has measure $2^{-n}$ (if you haven't learnt Lebesgue measure before, it's just the length of $I_n$). Their union $U$ is an open set containing all rationals, but has ...


1

It seems like you have two questions, one of which I addressed in a comment. The other is how to justify the statement Since $\lim_{n\to \infty} f_n(x)=f(x)$ for all $x\in M$, it follows that $\cap_{n=1}^{\infty} F_n =\emptyset$ Note that $$ \bigcap_{n=1}^\infty F_n = \left\{x \in M : |f_n(x) - f(x)| \geq \epsilon \text{ for all } n\right\} $$ ...


1

the result is correct but in the demonstration it is best not to confuse the symbol $\varepsilon$ and $\delta$ Since $f$ is uniformly continuous then $\forall\epsilon>0, \exists\delta>0$ s.t. $\forall x,x'\in X, d(x,x')<\delta\Rightarrow\rho(f(x),f(x'))<\epsilon$ But since $(x_n)$ is Cauchy in $X$, then $\forall\delta>0, \exists N$ s.t. $\...


0

I don't intend to write a completely new proof because your argument looks fine to me but I think that you can write the proof in a more systematic way. Although I think that your essential idea is correct, it is better to emphasize a few subtleties in the writing of the proof. Since $f$ is uniformly continuous then $$∀ϵ>0,∃δ>0\mid ∀x,x′∈X,d(x,x′)&...


1

In topological spaces, a cell complex is usually the same thing as a CW-complex. The name "cell complex" comes from the fact that there exists generalizations to other categories, but if you're interested in topological spaces then for all intents and purposes "cell complex" = "CW-complex". A finite cell complex is a cell complex that has a finite number of ...


1

De Morgan's laws if you want to be specific: $$M \setminus (\cup_{i \in I} A_i) = \cap_{i \in I} (M \setminus A_i) $$ $$ M \setminus (\cap_{i \in I} A_i) = \cup_{i \in I} (M \setminus A_i) $$ when $A_i, i \in I$ are subsets of $M$. The first one with $A_\lambda$ for $A_i$ for covering $A_\lambda$ is $\emptyset$ on the left hand side, and so the ...


1

You just rediscovered Cantor's original definition. A good discussion is here. Its wording is slightly different from yours: $$\forall m \in \mathbb{N}\ \lim_{n\rightarrow \infty} \left (x_{n+m}-x_n \right ) = 0 $$ And I think that this definition is more intuitive. You can give the following justification. You can start with the following exercise: ...


1

For $d(x,y)\ne 0$, we have $$\frac{1}{{d\left( {x,y} \right)}} = \frac{{1 + \left\| {x - y} \right\|}} {{\left\| {x - y} \right\|}} = 1 + \frac{1}{{\left\| {x - y} \right\|}} \geqslant 1 + \frac{1}{{\left\| {x - z} \right\| + \left\| {z - y} \right\|}} = \frac{{1 + \left\| {x - z} \right\| + \left\| {z - y} \right\|}} {{\left\| {x - z} \right\| + \left\| {z -...


0

Besides other great answers and references, I tried to follow your thoughts to complete the proof. \begin{aligned} d(x,z)+d(z,y) &= \frac{||x-z ||}{1+|x-z||}+\frac{||z-y ||}{1+||z-y||} \\ & = \frac{||x-z ||}{1+|x-z||} * \frac{1+||z-y||}{1+||z-y||} +\frac{||z-y ||}{1+||z-y||} * \frac{1+|x-z||}{1+|x-z||} \\&=\frac{||x-...


1

Changing the metric $d$ to $F(d)$ preserves the properties of a metric if: $F(0) = 0$ $F(d) > 0$ when $d > 0$ $F(a+b) \leq F(a) + F(b)$ for all $a,b \geq 0$ This is implied by, and in practice is equivalent to, $F$ being an increasing concave function. It is possible to artificially construct examples of $F$ that are increasing, subadditive and ...


1

Notice that the function $a\mapsto a/(1+a)$ is increasing and use that $d'(x,y)=||x-y||$ is also a metric (satisfies triangle inequality).


1

You must be talking about bounded linear operators, not limited. In normed vector spaces it holds that a linear operator is bounded if and only if it is continuous. This even is given as a definition sometimes. Now, if your definition of bounded linear operator is that it maps bounded sets to bounded sets then: From your expression for $||A||$ you get that $...


1

On some level this proof will be a proof by cases. However, using cases based on the trichotomy law $s<r$ or $s>r$ or $s=r$ are going to cause you headaches, as they already seem to have done for proving the separation axiom $d(x,y)=0$ iff $x=y$. First, don't just stick with that cumbersome formula for $d(x,y)$. Simplify it, by making good use of ...


2

I will try to address this in somewhat detailed form and using elementary methods. First, let us map the square $S$ (possibly with some boundary points removed) homeomorphically onto the disk $D$ (with corresponding boundary points removed). For convenience, I will take $S = [-1,1] \times [-1,1]$ and $D$ = unit disk, which is inscribed in $S$. We don't ...


1

They're homeomorphic since both of them are homeomorphic to the closed unit disc in $R^2$ with an arc removed on its boundary.


2

I think it should be métrica geodésica.


1

The function $f$ is not defined over $\overline{B}$, but it is over $A\cap\overline{B}$. However, your argument can be easily fixed. Take $x\in A\cap\overline{B}$, which exists by assumption. Since you're in a metric space, there is a sequence $(x_n)$ in $B$ converging to $x$. If $a$ is the common value of $f$ on $A$ and $b$ is the common value of $f$ on $...


1

You’re looking in the right general direction, but you’ve pretty much got the argument organized back to front. Starting with a continuous function $f:A\cup B\to\{0,1\}$ is fine; you know that $f$ is constant on $A$ and on $B$, and you want to show that it must be constant on $A\cup B$. In other words, you know that there are $a,b\in\{0,1\}$ such that $f(x)=...


0

I think what's missing in your proof is showing that, if $X$ is not connected, there is some continuous function that maps $X$ to a two point discrete space (hint: what can you say of the connected components of $X$?).


0

You are on the right track, but perhaps you should explain how your answer relates to functions from $X$ to the reals. If $X=A\cup B$ where $A, B$ are each open and non-empty with $A\cap B=\emptyset$ , let $f(x)=0$ for $x\in A$ and $f(x)=1$ for $x\in B.$ If $U$ is an open subset of the reals then $f^{-1}U$ is one of $\emptyset, A,B,X$, so $f^{-1}U$ is ...


1

The basic construction of "adding" two paths: suppose $f: [0,1] \rightarrow X$ is a path from $a$ to $b$, and $g: [0,1] \rightarrow X$ is a path from $b$ to $c$. So these are continuous maps with $f(0) = a, f(1) = b, g(0) = b, g(1) = c$. Then define $h: [0,1] \rightarrow X$ by $h(t) = f(2t)$ when $0 \le t \le \frac{1}{2}$ and $h(t) = g(2t-1)$ for $\frac{1}{...


0

Suppose $x\in A_i$ and $y\in A_j$. We may assume without loss of generality that $i<j$. Choose points $x_k\in A_k\cap A_{k+1}$ for $i\leq k<j$. Concatenate paths from $x$ to $x_i$, $x_i$ to $x_{i+1}$, etc., and finally from $x_{j-1}$ to $y$. Since this is probably for an assignment, you may want to phrase this as a proof by induction on $j-i$.


1

Given a subset $E$ of a (pseudo-)metric space $X$, define $\rho_E(x) = \inf\{d(x,y):y\in E\}$. I hope you know that $\rho_E(x) = 0$ iff $x\in \bar{E}$. Given your problem, simply set $$ f(x) = \frac{\rho_A(x)}{\rho_A(x)+\rho_B(x)}. $$


3

This map should do: $$(x,y)\mapsto\begin{cases}(x+\operatorname{sgn}(x)(|y|-1),y),&|x|>1\\(xy,y),&|x|\le1\end{cases}$$


1

A constant sequence is Cauchy and convergent. Any such metric space would have to be a null set.


0

No. Every constant sequence is a Cauchy sequence.



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