# Tag Info

## Hot answers tagged general-topology

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 ...

5

The rationals and the irrationals will do.

4

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

3

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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