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5

HINT: For (a) let $x$ and $y$ be distinct points in $[0,1]$, let $U$ be an open nbhd of $x$, and let $V$ be an open nbhd of $y$. What do you know about the sets $[0,1]\setminus U$ and $[0,1]\setminus V$? How big is the union of two countable sets? Is $\Bbb R$ countable? For (b), start by showing that a sequence $\langle x_n:n\in\Bbb N\rangle$ in ...


4

Easy example: $X = (0,1), Y = (1,2)$. Their intersection is empty, but the intersection of their closures is not.


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No, these are always the same. More precisely, $\pi_0(X,x_0)$ is the set of homotopy classes of basepoint-preserving maps $S^0\to X$: that is, maps that send $1$ to $x_0$. Two such maps $f,g:S^0\to X$ are homotopic iff there is a path in $X$ from $f(-1)$ to $g(-1)$, since given any such path you can define a homotopy $H:S^0\times I\to X$ to be the constant ...


3

Each set on the right is a subset of $\mathbb{R}$, so it is enough to show that every real number is in one of the three sets on the right. Fix $x\in \mathbb{R}$. If $x$ is in the interior of $X$ or of $X^c$, then we are done. Otherwise $x$ is not in the interior of $X$, so every interval centered at $x$ contains points in $X^c$, and $x$ is also not in the ...


2

Conventionally, in a metric space $(X,d)$ (in your case it's just the complex plane with Euclidean metric), the distance between any two nonempty sets $A,B$ can be defined as $$d(A,B):=\inf_{x\in A,y\in B} d(x,y).$$ (The infimum is of course defined because of the non-emptiness.) [EDIT: Also note that the inf is not necessarily attained in general. A quick ...


2

Using your definitions, this is a lot easier than the linked proof. Let $x \in \mathbb{R}.$ Then one of three cases is possible: (i) some open neighborhood of $x$ is contains only points in $X$ (ii) some open neighborhood of $x$ is contains only points in $\mathbb{R}-X$ (iii) none of the above If (i) then $x \in \mathrm{int}(X)$; if (ii) then $x \in ...


2

There are various papers from a few decades ago which compute the Hausdorff dimension of the complement of $U$ in various Apollonian gasket kinds of constructions. If the Hausdorff dimension of the complement of $U$ is $<2$ then that's sufficient to conclude that the Lebesgue measure of the complement of $U$ is zero, and so then yes, $U$ has full Lebesgue ...


2

Here is an example, but to understand it you need some complex analysis. Let $U$ be the domain in the complex plane bounded by the Warsaw circle $W$: Then $U$ is bounded and simply connected. Let $f: D\to U$ denote the Riemann mapping, i.e. the (essentially) unique conformal mapping. Then $f$ is a homeomorphism to its image. However, $f$ does not admit ...


2

Countable basis $\Longrightarrow$ separable | in general topological spaces. Separable $\Longrightarrow$ countable basis | only in metrizable topological spaces. Sorgenfrey’s plane is not metrizable. It is separable, but does not admit a countable basis.


2

The closure of a Baire space is Baire is short for the following theorem: Suppose $D$ is dense in $(X,\mathcal{T})$ and suppose that $D$ is Baire in the subspace topology. Then $X$ is Baire. Proof: let $U_n \subseteq X$ be open and dense. Then consider $V_n = D \cap U_n$ for all $n$. The $V_n$ are by definition open in $D$, they are non-empty as $D$ is ...


2

Here is an answer, but please read what follows the proof. If $\tau$ is the co-finite topology on $X$ than if $U\in \tau$ then $X-U$ is finite. Now if $\tau$ is also the discrete topology then for every $x\in X$ we have that $\{x\}\in\tau $ , So $\{x\}$ is co-finte. i.e $X-\{x\}$ is finite. And from here it should be clear which option is true. You are ...


1

Suppose that all distinct $x$ and $y$ have disjoint neighbourhoods, and suppose that $(x_i)_{i \in I}$ is a net converging to both $x$ and $y$. By assumption on $X$ find open $U$ with $x \in U$ and open $V$ with $y \in V$ such that $U \cap V = \emptyset$. Now apply the definition of convergence to $x$ and $U$. So there exists $i_0 \in I$ such that ... ...


1

strictly speaking a topological space is a pair consisting of a set $X$ and a set $\tau\in \mathcal{P}(\mathcal{P}(X))$ of "open subsets of $X$" (of course satisfying some properties). Thus the correct way to adress a top. space is to write down a pair $(X,\tau)$. However, as people are lazy, they just write $X$ or $\tau$ whenever they feel that the ...


1

As noted in the comments, this is false. Intuitively, on the left-hand side in 1) you add one point to both spaces and then take the cartesian product whereas on the right-hand side you first take the product and then add one point. For example let $A=B=C_0((0,1))$. Then $\widetilde{A\otimes B}=\widetilde{C_0((0,1)\times (0,1))}=C(S^2)$ and $\tilde A\otimes ...



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