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5

Consult your definitions—the interior of a set is its maximal open subset. If your topology is e.g. discrete, then $\{ x \}$ is open, so it is its own interior. But in the standard topology on the reals, it's not open. In this case, the empty set is its only open subset, so that is its interior. Both of the above cases are metrisable, so the restriction to ...

3

Imagine that you place a disk on top of a sphere so that the center of the disk is at the north pole. Then spread the disk over the surface of the sphere, and glue the boundary of the disk at the south pole. This is one way to see the homeomorphism that the book is talking about. To construct such a homeomorphism explicitly, we can use polar coordinates ...

3

The space resulting from gluing opposite edges of an octagon naturally depends on the orientation of the gluing, but assuming "antipodal edges" are glued with the same orientation, as when building a torus from a square, the result is indeed a two-holed torus. It's easy to see that Two vertices that are endpoints of a dashed segment are identified, so ...

2

I wanted to tell you that you better let $N$ be the $\max$ instad of the $\min$ of $N_1,N_2$. But actually, we can get along with $N=N_2$. Also note that you don't want to show anything about $|a_m-a_n|$. Instead you want to show that $|a_n-L|<\epsilon$ (to conclude $a_n\to L$; and incidentally you use both $a$ and $L$, both intended to mean the same): ...

2

The metric $d_2$ needs to be defined on the whole space. Are you taking $\arctan(\infty) = \frac \pi 2$? I'll assume that's the case. In order for $\{\infty\}$ to be open, there must be $\epsilon > 0$ satisfying $B(\infty,\epsilon) \subset \{\infty\}$ However, for every $\epsilon > 0$ you have B(\infty,\epsilon) = \{b \in X : d_x(\infty,b) < ... 2 A path in \mathbb{R}^n is a continuous function f from some interval [a,b] to \mathbb{R}^n. We can also not require the interval to be closed. Think of the image of any continuous function on [a,b]. Does it not look like you're moving on a path from f(a) to f(b) in the image, without "jumping" across points, as you go from a to b in the ... 2 You can see that that \partial(X\times Y) \neq \partial X\times\partial Y by considering X = Y = (0, 1) \subset \mathbb{R}. On the one hand, \partial(X\times Y) is the boundary of the unit square, while \partial X\times \partial Y consists of the four corner points. The relationship between the boundary of a product and the boundaries of the ... 2 Assume that H\subset G_0 is an open subgroup of G contained in the connected component of the identity G_0. Then write G_0 as a disjoint union of cosets G_0 = \bigsqcup_{g} (g+H), for some set of coset representatives. Since H is open, so is each coset g+H. Now if H\subsetneq G_0, then G_0 = H \sqcup \bigsqcup_{g\neq 0} (g+H) ... 1 Hint Any ball contains a point with all coordinates rational, and \mathbb Q^n is countable. 1 The idea of a path (to me) is just a 1-dimensional curve that's parameterized. For example, if we walk around the top half of the unit circle, we can parameterize this path by \begin{align*}\vec{x}: [0, \pi] &\to \Bbb R^2 \\ t &\mapsto \left(\cos t, \sin t\right). \end{align*} I believe the notation C^k just stands for the class of functions ... 1 The answer to your question is "Yes, \{x\in X;\;a\leq x\leq b\} for a<b can be an open set." There are a number of ways this can happen. The trivial but slightly boring one is if your order has endpoints a, and b such that \forall x\in X\; a\leq x and \forall x\in X\; b\geq x. Then X=\{x;a\leq x\leq b\} and is open. An example of this is ... 1 Define N\ge max(N_1, N_2) as an integer such that n_w\ge N_2, \forall w\ge N. Then\mid a_m-L\mid\le\mid a_m-a_{n_m}\mid+\mid a_{n_m}-L\mid\lt\epsilon, \forall m \ge N. Hope this helps.

1

I just answered this question for a homework assignment, so I've copied my solution below. For each $x \in X\setminus A$, define $U_x := X\setminus\{x\}$. Then $\{x\}$ closed $\implies U_x$ open. Moreover, $\overline{U}_x = X$ as $\{x\}$ is not an open since $x \notin A$. Since $X$ is countable, we have \begin{equation*} \bigcap\limits_{x \in X ...

1

Basically, what you want to do is tho glue two torus. Both torus can be identified with the following representation: Now to glue them together, you want to cut a small piece of each torus, and glue them in a certain way on the area you cut. We are going to cut a small triangle at points $p$ and $q$ as following way: Note that $p$ and $q$ are note at ...

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