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I'm assuming a normed space $X$ based on your comments. If $X$ is an infinite-dimensional normed space, then there exists a sequence $\{ x_{n}\}_{n=1}^{\infty}$ of unit vectors such that $\|x_{m}-x_{n}\| \ge 1/2$ for all $n \ne m$. That's a consequence of the Riesz lemma. I think that answers your question.


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You can repeat the "sines and cosines" process one more time if you want. Like our friend @Bruno Joyal said: For the 1-sphere: $X(\phi)=(\cos(\phi),\sin(\phi))$, so that its squared coordinates sums up to 1. $\cos^2(\phi) + \sin^2(\phi) = 1$ For the 2-sphere: $X(\phi,\psi)=(\cos(\phi)\cdot\cos(\psi),\sin(\phi)\cdot \cos(\psi),\sin(\psi))$, so that its ...


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Many important quotient spaces are not metric, nor even Hausdorff. This happens quite commonly for orbit spaces of group actions, which are very important in geometric group theory, in dynamical systems, etc.


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You are misunderstanding the term "unbounded". This term means that the set contains elements of arbitrarily large magnitude, i.e., it is not contained in $[-N,N]$ for any positive number $N$. Your comment suggests you are thinking of the term as meaning "not containing its boundary". This is incorrect. The set $(0,1)$ which you cite is indeed bounded (and ...


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Your proof idea works, but the phrasing is imprecise and crucial details are left out. What you're trying to do is to show that for any finite subcollection $O_{\lambda_1},\ldots O_{\lambda_n}$,the following holds: $A\cap K\subsetneq O_{\lambda_1}\cup\ldots \cup O_{\lambda_n}$ or $B\cap K\subsetneq O_{\lambda_1}\cup\ldots \cup O_{\lambda_n}$. This amounts to ...


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It is because $\mathbb C = \mathbb R ^2$ is locally path connected, and so every open subset of $\mathbb C$. For a locally path connected topological space, the path-connected components are open and coincide with the connected components. In fact, let $X$ be a topological space which is locally path connected and $C\subset X$ a path connected. If $p \in ...


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Let $\Omega_p$ be the path component of $\Omega$ which contains the point $p$ and let $x$ be any point in $\Omega_p$. Let $\epsilon$ be such that $B_{\epsilon}(x)\subset\Omega$ which must exists because $\Omega$ is open. Note that $B_{\epsilon}(x)$ is an open disk and so is path connected. In particular for any $y\in B_{\epsilon}(x)$, the element $y$ must be ...


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Let $P$ be any path component of $\Omega$. Let $p \in P$ be any point. Since $\Omega$ is open, there exists an open ball $B$ such that $p \in B$ and $B \subset \Omega$. Since $B$ is path connected, it must be contained in $P$, by definition of path component. Therefore $P$ is open.


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Maybe this is helpful. Let it be that $f:X\rightarrow Y$ is continuous at each element $x\in X$. If $V$ is open in $Y$ then for each $x\in X$ with $f(x)\in V$ there is an open set $U_x\subset X$ with $x\in U_x$ and $f(U_x)\subset V$ or equivalently $U_x\subset f^{-1}(V)$. As union of open sets $\bigcup_{x\in f^{-1}\left(V\right)}U_{x}$ is open in $X$ ...


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As Mark Bennet said, it means every two regions have a nontrivial piece of boundary in common: nontrivial means containing a topological arc, i.e., not being a point. This is a standard detail in Map coloring problems, and Thurston's exercise is in the context of map coloring.


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What's wrong is the step where you say $X\setminus\prod_{i=1}^\infty C_i=\prod_{i=1}^\infty(X_i\setminus C_i)$. Not so. For example, suppose $X_i=\{0,1\}$ and $C_i=\{0\}$. What is $X\setminus\prod_{i=1}^\infty C_i$ and what is $\prod_{i=1}^\infty(X_i\setminus C_i)$? Hint: $\prod_{i=1}^\infty C_i=\bigcap_{i=1}^\infty\pi_i^{-1}(C_i)$. Use the facts that the ...


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Uncountable additivity are considered especially in set theory. For example, an uncountable cardinal $\kappa$ is a real-valued measurable cardinal if and only if there exists a nontrivial $\kappa$-additive probability measure on $\kappa$. Although in some cases $\sigma$-additivity is enough. For example, the least cardinal which has a $\sigma$-additive ...



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