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5

Hint: It's usually way easier to show path connected than connected. Path connected implies connected, it's stronger. So take two arbitrary points and show you can connect them with a path. There are two cases: The points are not on the same line as the origin on opposite sides (The easy case), and they are: (The almost as easy case) Can you manage?


4

Your "formula" for $\mathbb{R}^n\backslash \{0\}$ is wrong. Draw $\mathbb{R}^2$ and you will understand why. If you want to avoid usage of path-connectedness, do the following. For $n>1$, take $A,B \subset \mathbb{R}^n \backslash \{0\}$ given by $A:=\{x \in \mathbb{R}^n \mid x_n > 0\}$ and $B:=\{x \in \mathbb{R}^n \mid x_n<0\}$. $A$ and $B$ are ...


3

Note that the function $f(A) = det(I - A)$ is continuous. So $U = f^{-1}(\mathbb{R} - \{0\})$ is open as a preimage of an open set under a continous map.


2

The idea of the proof is fine, but you some small mistakes, note that $$ f^{-1}\bigl((-1,1)\bigr) = \color{red}{(-\infty,} 0] \cup {\color{red}(}1,\infty) $$ which is also not an open set. No, this would not work for any neighbourhood, due to the fact that $f$ continuous at $0$ means $$ \forall U: \bigl(U \text{ open neighbourhood of $0$}\bigr) \to \bigl(f^...


2

Your questions reveal a typo that Valette was not aware of. The third line of the proof of Proposition 3.3.7 should be:$$\pi_n(\text{GL}_\infty(A)) = \pi_{n - 1}(\Omega \text{GL}_\infty(A)) = \pi_{n - 1}(\text{GL}_\infty(SA)),$$where $\Omega X$ is the loop space of $X$. The first equality is a basic property of loop spaces, see e.g. here. The second follows ...


2

No, at least in the sense that there are no such functors $F$ which are substantially more approachable than the original problem. There are very few general techniques for distinguishing homotopy equivalent, non-homeomorphic spaces: see the following link for work in the last twelve years distinguishing some such pairs of 3-dimensional closed manifolds, ...


2

Restrict attention to base elements. A base for the product topology is furnished by elements of the form $B=I_1\times\cdots \times I_N$ with each $I_k$ open. Each $f_k^{-1} (I_k)$ is open and their intersection which equals $f^{-1} (B)$ is thus open. The statement is true also in infinite dimension (small modification of proof since you should only take ...


2

It does not matter whether you consider the size of a minimal base or a subbase because they are the same (for infinite bases/subbases): if we have a subbase of size $\kappa$, its generated base (using finite intersections) is also of size $\kappa$. The weight $w(X)$ of a space $X$ is defined as the minimal size of a base for $X$. So you are asking for ...


1

The map $f$ is proper as any compact subset of $\mathbb{D}$ is closed (in the topology of $\mathbb{C}$ and not in the subspace topology), and preimages of closed sets are closed, and closed subsets are closed, as the map $z \mapsto z^2$ on $\mathbb{C}$ restricts to $\mathbb{D}$. For $g$ take the closed compact ball $B_2(0)$ of radius 2. The preimage of ...


1

There are many different processes that let you go from a 3-dimensional polytope to an "analogous" 4-dimensional polytope. To get from a 3-simplex to a 4-simplex you take a join with a point outside the affine span of the 3-simplex. To get from a 3-cube to a 4-cube you take the product with a unit interval. To get from 3-dimensional octahedron to the 4-...


1

HINT: You’re right to worry: an interval in a linearly ordered space can be both closed and open. Your interval is neither: it’s open at the left end and closed at the right end. In interval notation it’s $$\left(\left\langle\frac12,\frac12\right\rangle,\left\langle\frac12,1\right\rangle\right]\;.$$ The way to show that this one is not open is to find a ...


1

It is often very difficult to calculate. A point $y\in X$ is in the weak closure if you can not enclose it in a weak neighborhood disjoint from $S$, i.e. if for every $\epsilon>0$ and linear functionals $\ell_1,\ldots,\ell_k\in X'$ the set $N=\{ x\in X : |\ell_i(y-x)|<\epsilon, i=1,\ldots,k \}$ intersects $S$. Note that $k$ must be finite. In finite ...



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