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3

Hint Define a sequence of polynomials by $$ P_n(x)= \frac{1}{\log(\log(n))}\sum_{k=0}^n \frac 1{k+1}x^k $$


2

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), and $$Q\subseteq(\pi_1(q)\times \pi_2(Q)).$$ 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 ...



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