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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 ...

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