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Hint Define a sequence of polynomials by $$P_n(x)= \frac{1}{\log(\log(n))}\sum_{k=0}^n \frac 1{k+1}x^k$$
For a given $p\gt1$, consider $$a_{n,k}=\left\{\begin{array}{cl} \dfrac1{n^{1/p}}&\text{if }1\le k\le n\\ 0&\text{if }k\gt n \end{array}\right.$$ Then \begin{align} \left(\sum_{k=1}^\infty a_{n,k}^p\right)^{1/p} &=\left(\sum_{k=1}^n\frac1n\right)^{1/p}\\[6pt] &=1 \end{align} while \begin{align} \sum_{k=1}^\infty a_{n,k} ... 2 Yes it does. From That definition we are are talking about all points within a radius r of p. 2 The set is convex. Take any two points a and b in the unit square, consider the line joining them a+t(b-a) where t\in[0,1], and check that the components of every point on the line lie in [0,1]. 1 Your definition of convexity only applies to subsets of the real numbers. In any higher-dimensional space the correct definition is that for all a, b \in Y and \lambda \in [0, 1] the vector \lambda a + (1 - \lambda) b must lie in Y. You can easily verify directly that any cartesian product of convex sets is again convex. Also note that the ... 1 In point 2, you don't need to mention that \pi_{\alpha} is open. In point 3, you mean that U, V are open. There is a major error in point 4, because \pi_{\alpha}(U) and \pi_{\alpha}(V) need not be disjoint. (Also, \pi_{\alpha} being open has nothing to do with the inclusions A_{\alpha} \subseteq \pi_{\alpha}(U) and B_{\alpha} \subseteq ... 1 The sets (0,1)\cap \Bbb{R}\backslash \Bbb{Q} and [0,1] \cap \Bbb{R}\backslash \Bbb{Q} are respectively open and closed in \Bbb{R}\backslash \Bbb{Q} by definition of the subspace topology. But they are both equal to [0,1] \backslash \Bbb{Q}. So [0,1] \backslash \Bbb{Q} is both open and closed in \Bbb{R}\backslash \Bbb{Q}. Therefore, it is it's own ... 1 You are partially correct in your thinking: your set is open, but, it is also closed. Since [0,1] \setminus \mathbb{Q} = (0,1) \setminus \mathbb{Q} = (0,1) \cap (\mathbb{R} \setminus \mathbb{Q}), and since the intersection of an open set with a subset is open in the subset topology, your set is closed in \mathbb{R} \setminus \mathbb{Q}. On the other ... 1 Supppose Q is compact in G_1\times G_2. then \pi_1(Q) is compact, \pi_2(Q) is compact (the \pi_i are the canonical projections), andQ\subseteq(\pi_1(q)\times \pi_2(Q)).$$It is also a closed subset of these provided the factor spaces are Hausdorff. Conversely a closed subset of a product of two compact spaces is compact. So you have this ... 1 Partial answer, for p>2. Let P_n(x)=\sum_{j=1}^n j x^j . We have$$\|P_n\|^p=\sum_{j=1}^n j^p<\sum_{j=1} ^n\int_j^{j+1}y^p dy=\int_1^{n+1}y^p dy==\frac{(n+1)^{1+p}-1}{1+p}<\frac {(n+1)^{1+p} }{1+p}.\text {So }\; \|P_n\|<\frac {(1+n)^{(1+1/p)}}{(1+p)^{1/p}}\;\text { But }\; P_n(1)=(n^2-n)/2.$$Let Q_n=P_n/n^{1+2/p}.\; Then ... 1 As Willard says, this is a little tricky. The key idea can be seen in my comment to Hagen von Eitzen above. HINT: First prove the following lemma. Lemma. Let \langle X,d\rangle be a metric space, let U be a non-empty open set in X, and let \epsilon>0. Then there is a discrete set D_\epsilon(U)\subseteq U such that \bigcup_{x\in ... 1 Let X be the middle-thirds Cantor set with the Euclidean metric, and let x=\epsilon=\frac13. Then$$B_\epsilon(x)=\left(0,\frac23\right)\cap X=\left(0,\frac13\right]\cap X\;,$$whose boundary in X is \{0\}. Let$$V=\left(0,\frac29\right)\cap X=\left(0,\frac19\right]\cap X\;; then $V$ is open in $X$, $V\subsetneqq B_\epsilon(x)$, and ...