Tag Info

Hot answers tagged

4

Let's continue your proof. Let $x_0$ as in your proof. Define the sequence $x_n=f(x_{n-1})$. The following holds: $$ d(x_{n+m},x_n) \ge d(x_{n+m-1},x_{n-1}) \ge .... \ge d(x_m,x_0) > \delta $$ In particular, the sequence $x_n$ has no convergent subsequence. Contradiction because $X$ is compact.


4

Your argument is correct and is probably the most straightforward. The result does not even hold in all $T_3$ spaces: in this answer I give a general method, due to Eric van Douwen, for starting with a $T_3$ space having two points that cannot be separated by a continuous real-valued function and producing from it a $T_3$ space on which all continuous ...


3

The proof seems fine but Tietze needs to have the function continuous on a closed subset, but as it has no limit points this is given. If your function Needs to be continuous I can tell you the third question to have a negative answer. In fact there is a non compact Hausdorff space such that the continuous functions are exactly the constant ones. Take ...


3

Your reasoning is correct. Think about $\mathbb R^2$ with the usual euclidean metric. An open ball does not contain its boundary. In general, the open ball $B_r(x_0)$ in a metric space $(X,d)$ is defined to be $$B_r(x_0) := \{x \in X : d(x_0,x)<r\}$$


3

Try to prove that $x, f(x), f(f(x)), f(f(f(x))),\ldots$ is a Cauchy sequence. Let $a$ be its limit. Then see if you can show that for every $\varepsilon>0$, $d(a,f(a))<\varepsilon$. A function $f$ for which there exists $K$ between $0$ and $1$ such that for all points $x,y$ one has $d(f(x),f(y))\le Kd(x,y)$ is called a contraction.


3

Suppose $X$ is not bounded. Fix $x\in X.$ Claim: The collection of $d(x,y),~y\in X$ is not bounded. If it were bounded by some $C,$ then for any $y,z\in X$ we would have \begin{equation*} d(y,z)\leq d(y,x)+d(x,z)\leq 2C \end{equation*} which is a contradiction. Since the definition of sequential compactness needs every infinite sequence to have a ...


3

In this case no such solution exists. Let $d_E$ be the Euclidean distance. As you note, if $d(x, \cdot)$ is $d_E$-continuous for each $x$ then $(S^2, d)$ will be a coarser topology than $(S^2, d_E)$. So the identity map of $S^2$, considered as a map from $(S^2, d_E)$ to $(S^2, d)$ is a continuous bijection from a compact space to a Hausdorff space; any ...


2

The statement is of course not true. Take $G$ to be the entire space, for example. But what you can prove is that there is some $n$ such that $G\cap K_n$ is not empty, and therefore relatively open there. And then it meets the relevant dense set. (Of course, assuming $G\neq\varnothing$, which is of course the initial assumption.)


2

You are right. By the very definition of $\def\eps{\varepsilon}B_\eps(x)$, we have $$ B_\eps(x) = \{y \in X : d(x,y) < \eps\} $$ hence $$ B_1(x) = \{y \in X : d(x,y) < 1 \} $$ Now the points with distance $1$ to $x$ (that is all points but $x$), do not belong to $B_1(x)$.


2

Take $U=\{e_n|\:n\in\mathbb N\}\subset \ell^\infty (\mathbb R)$. It is obviously bounded since $\forall x\in U \:\|x\|=1$, but $\forall x,y\in \ell^\infty$ we have $d(x,y)=1$, so obviously for $\epsilon=1$ there is no finite number of open balls with radius $\epsilon$ that cover $U$ - cause each ball would contain at most one member of $U$.


2

Take $X=\mathbb{R},\, Y=\{0\}$ and $x_{n}=n$ . $\pi$ defined by $\pi(x)=0$ for every $x\in\mathbb{R}$ Then $\pi$ is open and surjective but $x_{n}$ is not convergent while $\pi(x_{n})\equiv0$ is


2

I claim that if $\pi$ is injective we get the result Let $y\in Y$ be (a) limit of $\pi(x_n)$. Let $x=\pi^{-1}(y)$. Let $U\subset X$ be an open neighborhood of $x$. $\pi$ is open, and therefore $\pi(U)$ is a open neighborhood of $y$, so there is $N\in\mathbb N$ such that for every $n>N$ we have $\pi(x_n)\in \pi(U)$ so $x_n\in U$ for all $n>N$ as ...


2

As a partial answer, here's an example where you can't take all of the $A_n$ to be open. Essentially, the idea is to diagonalize against a countable sequence of moduli of continuity. Let $\Omega$ be the set of all continuous nondecreasing functions $\omega : [0,1] \to [0,1]$ with $\omega(0)=0$. Let $X = \Omega \times [0,1]$ with the metric $$d((\omega_1, ...


2

I thank Nate Eldredge and Alex Ravsky for contributing. I think I can now conclude that the property I'm looking for cannot be guaranteed in general. To see this, let $X$ be any infinite-dimensional separable Banach space—for example, $L^p(\mathbb R)$ for any $p\in[1,\infty)$. A theorem by Izzo (1994) I mentioned in a comment above guarantees the existence ...


1

Let $Q$ be the space of non-negative rational numbers, $N$ the natural numbers including $0$, and $q:Q\to Y=Q/N$ the quotient map identifying $N$ to a single point. Note that $Y$ is a Hausdorff normal space since $Q$ is such a space and $q$ is a closed map. Now look at $Y\times Q$. This space is Hausdorff. We will show that it's not compactly generated. Let ...


1

Sorry I mislead you in the comments: Converse: Suppose $B_r(x)\cap E^c\ne\emptyset \forall r>0$. $x\notin E^0$ because $E^0$ is open (and hence we can find an open ball about x such that $B_r(x)\subset E^0$). Edit I would prefer to write it like this; Converse: Suppose $B_r(x)\cap E^c\ne\emptyset \forall r>0$. Suppose $x\in E^0$. Since $E^0$ ...


1

Define a new metric $d$ on $\Bbb R$ by $d(x,y)=\min\{|x-y|,1\}$; you can easily check that $d$ generates the usual topology on $\Bbb R$. Every subset of $\Bbb R$ is bounded with respect to $d$, so we need only find a subset that is not totally bounded. $\Bbb N$ will do: if $F\subseteq\Bbb N$, then $$\Bbb N\cap\bigcup_{x\in F}B_d\left(x,\frac12\right)=F\;,$$ ...


1

Recall that the topology on $\mathbf{R}$ is generated by finite intervals $(a,b)$, meaning an open set is any set you may form through finite intersection and arbitrary unions of finite intervals $(a,b)$. The second set is closed. To see this, consider its complement, $(-\infty,0)\cup (0,2)\cup(2,\infty)$, which is a union of open sets: a finite open ...


1

Since I can't comment (low rep), the interior of $A=[0,1]\cup\{2\}$ would be $(0,1)$. Take any $x$ inside $\operatorname{int}A$. $x$ cannot be $0$, $1$ or $2$ since there isn't any $ε>0$ so that $(x-ε,x+ε)$ is fully inside $A$. So $x$ belongs to $(0,1)$ and so $\operatorname{int}A$ is inside $(0,1)$. And ...


1

If possible let us assume that $U_1$ is open and since $m\in U_1$ so there exists an $\epsilon>0$ such that $(m-\epsilon,m+\epsilon) \subset U_1$. Now there are two possibilities, either $q \in (m-\epsilon,m+\epsilon)$ or $q\notin (m-\epsilon,m+\epsilon)$. But you can clearly see that the only possibility is actually $q \notin (m-\epsilon,m+\epsilon)$ ...


1

Note that if every subset is open, then every subset is closed: Given $A \subset X$, then the complement $A^c = X \setminus A$ is a subset, therefore open, and $A^c$ open is equivalent to $A$ is closed. If you want to be concrete, you can view the complement of a single point as the union of the balls of radius $1/2$ centered on $y$, as $y$ ranges across ...


1

No, it's not fine. $A\cup B$ is the union of the intervals, not their intersection. $$A\cup B = [-3;2]\cup[1;4] = [-3; 4] \\[1ex] A\cap B = [-3;2]\cap[1;4] = [1; 2] $$ Hint: If $(A\cup B)^\circ\neq A^\circ\cup B^\circ$ then you need to select two sets so that there exists some element that is in the interior of the union but not in the union of the ...


1

To show that every A-space satisfies the given property, let $(X,d)$ is an A-space, let $F \subseteq X$ be closed. Consider the quotient space $Y = X / F$ and the natural quotient mapping $q : X \to Y$. Let $* \in Y$ denote the point corresponding to the collapsed closed set $F$. Note that $Y$ is clearly Hausdorff, and $q$ is a closed, continuous, onto ...


1

There is a problem with your contrapositive. Hint: show that a closed part of a compact set is compact.


1

Of all the properties of a distance, the one requiring some work is the triangle inequality. Before tackling it, note the following: in (3) take $\alpha = \frac 1 2, x_2=0, x_1=x$. Then $\phi (\frac 1 2 x) \geq \frac 1 2 \phi (x)$ (using also (1)). Since $\phi$ is increasing and $\rho$ satisfies the triangle inequality , $\phi \circ \rho (x,y) = \phi (\rho ...


1

You could prove that the metric $d$ is a continuous function from $A\times A$ to $\mathbb{R}$ by considering the product topology on $A\times A$. Let $M = \left({A, d}\right)$ be a metric space. Let $\tau$ be the topology on $A$ induced by $d$. Let $\left({A \times A, T}\right)$ be the topological product of $(A, \tau)$ and $(A, \tau)$. ...


1

Since $$ d:X\times X\to [0,\infty) $$ is a metric, it satisfies the triangle inequality: $$ d(a_1,a_2)\le d(a_1,a_3)+d(a_3,a_2) \quad \forall a_1,a_2,a_3\in X. $$ Given $a=(a_1,a_2)\in X\times X$, we have for every $x=(x_1,x_2)\in X\times X$: $$ d(x_1,x_2)\le d(x_1,a_1)+d(a_1,a_2)+d(a_2,x_2), $$ and therefore $$\tag{1} d(x_1,x_2)-d(a_1,a_2)\le ...



Only top voted, non community-wiki answers of a minimum length are eligible