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8

I think the result does not hold. Suppose Red is the closed unit disk (centered at the origin); blue is the closed annulus centered at the origin with inner radius $1$ and outer radius $2$ (so blue completely surrounds red); and green consists of all points at distance greater than or equal to $2$ from the origin.


8

It's not true. Consider $x=0$, $y=1$, $z=2$. Then $$4 = d(x,z) \not\leq d(x,y) + d(y,z) = 1 + 1 = 2$$


6

If blue has a connected boundary, then it holds. Because if there were no point of intersection of all three sets, you could split $\partial\mathrm{Blue}$ to $\partial\mathrm{Blue}\cap\mathrm{Red}$ and $\partial\mathrm{Blue}\cap\mathrm{Green}$ which would yield a decomposition of $\partial\mathrm{Blue}$ into two open and closed sets, contradicting its ...


4

Yes. Every countable ordinal can be realized as a subspace of the rationals, and therefore as a subspace of the real numbers. You can do this by transfinite induction, and note that both $\Bbb R$ and $\Bbb Q$ have the property that every open interval is homeomorphic, and order isomorphic, to the full space.


3

Every countable linearly ordered space with its order topology can be continuously embedded in $\Bbb Q$. Here’s a sketch of a proof. Let $\langle X,\preceq\rangle$ be a countable linear order. If $x,y\in X$, $x\prec y$, and $(x,y)=\varnothing$, write $y=x^+$ and $x=y^-$. Let $L=\{x\in X:x=y^-\text{ for some }y\in X\}$. For each $x\in L$ let ...


3

If all you want is a homeomorphic embedding of a countable ordinal $\alpha$ into $\mathbb R$, just take an injection $f:\alpha\to\omega$ and define $h:\alpha\to\mathbb R$ by setting $$h(\xi)=\sum_{\beta\lt\xi}2^{-f(\beta)}.$$


2

As an example, consider the closed unit disk $D = \{(x,y) \in \mathbb{R}^2: x^2 + y^2 \le 1\}$. This is a closed set when considered as a subset of $\mathbb{R}^2$, in its usual Euclidean metric. And it's not open there, as shown by points like $(1,0)$: every open ball around $(1,0)$ with radius $r$ contains a point $(1 + \frac{r}{2},0) \notin D$, so there is ...


2

You've got the right idea. First note that $$U = \{z \in D \mid |f(z) - 1| < 1\} = \{z \in D \mid f(z) - 1 \in \mathbb{D}\} = \{z \in D \mid g(z) \in \mathbb{D}\} = g^{-1}(\mathbb{D})$$ where $g(z) = f(z) - 1$. As $f$ is continuous, so is $g$, and as $\mathbb{D}$ is open, $U$ is open. Likewise, you can show that $V$ is open. The first thing you need ...


1

Let $W=\{z| |z^2-1| < 1 \}$. Since $|z^2-1|=|z-1||z+1|$, we see that $W = B(-1,1) \cup B(1,1)$. We are given that $f(D) \subset W$. We see that $\operatorname{re} B(-1,1) = (-2,0)$, $\operatorname{re} B(1,1) = (0,2)$, and so $\operatorname{re} W = (-2,0)\cup (0,2)$. Let $I = (\operatorname{re} \circ f) (D) \subset \operatorname{re} W = (-2,0)\cup ...


2

Your argument doesn’t need paracompactness: just show directly that every regular Lindelöf space is normal. This is fairly straightforward; see, for instance, Theorem $16.8$ in Willard’s General Topology. As you say, if $\Bbb R^I$ were Lindelöf, $\Bbb Z^I$ would also be Lindelöf, and I’ll show directly that it’s not. Let $\mathscr{P}=\{P\subseteq ...


1

Letting $\Bbb D(z_0;r)$ denote the open disk centered at $z_0$ of radius $r$, you have $U = f^{-1}(\Bbb D(1;1))$ and $V = f^{-1}(\Bbb D(-1;1))$. These sets are disjoint since $\Bbb D(1;1)$ is disjoint from $\Bbb D(-1;1)$. Since $f$ is continuous and the disks $\Bbb D(1;1)$ and $\Bbb D(-1;1)$ are open, $U$ and $V$ are open. So $\{U,V\}$ is a separation of $D$ ...


1

One of the $U$ and $V$ must be non empty. WLOG assume $U$ is non empty. We will show that $V$ is empty. Assume otherwise. Take a point $z_1$ in $U$ and a point $z_2$ in $V$. Then connect those two points by a path $p$. The function $g: p \rightarrow \mathbb{C}$ defined $ z \rightarrow |f(z)-1| $ is continuous. We have that $g(z_1)<1$ and $g(z_2)>1$ ...


1

Hint: Take $(X,d)=\ell^2$ endowed with the standard metric.


1

It seems the following. 3.For any monotonic function $f$ such that $f(1)\le 1/2$ and $f(n)\to 0$ and any irrational $\alpha\in [0,1]$ there exists an enumeration $(q_n)_{n\in \mathbb{N}}$ of the rationals in $[0,1]$ such that $U_{f,1}\not\ni\alpha$. For this purpose, while enumerating the rationals of $[0,1]$ it suffices to choose $q_n\ne (\alpha-f(n), ...


1

The idea of the covering dimension is due to Lebesgue, who claimed to have proven that $[0,1]^n$ has covering dimension $n$ (in modern terms; he formulated it in terms of properties of finite covers of sets of small diameter; Cech later gave a more formal definition of dimension based on the order (or ply) of a cover). Brouwer, somewhat later actually gave a ...



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