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5

I think it is much clearer to stipulate that $U$ is an open set. Expressing it as an element of the topology does not increase the clarity of what you are saying.


4

This is clear: the components of $f$ are continuous. Now use the universal property of product spaces.


3

Consider the subbasic open set $$S_{V,n}=\prod_{k\in\Bbb Z^+}U_k\;,$$ where $U_n=V$ is an open set in $\Bbb R$, and $U_k=\Bbb R$ for $k\ne n$. It’s not at all hard to calculate $f^{-1}[S_{V,n}]$: $f(t)\in S_{V,n}$ if and only if $nt\in V$. If we let $g:\Bbb R\to\Bbb R$ be the function $g(t)=nt$, we can express this in terms of $g$: $t\in f^{-1}[S_{V,n}]$ ...


2

Your proof is almost valid, but it is unfortunately circular, because you use the following lemma: $$a \text{ is odd} \implies a^{n-1} \text{ is odd}$$ However, if you take the contrapositive of this statement: $$a^{n-1} \text{ is even} \implies a \text{ is even}$$ which is basically the same as your statement, except with $n-1$ instead of $n$. Therefore, ...


2

The result is correct if $\sum_{i=1}^na_i$ is a multiple of $n$. Let $a=\frac1n\sum_{i=1}^na_i$; we want to show that $$\sum_{i=1}^n\binom{a_i}r\ge n\binom{a}r\;.\tag{1}$$ Suppose that $a_i<a<a_k$. Then $$\begin{align*} \binom{a_i}r+\binom{a_k}r&=\binom{a_i}r+\binom{a_k-1}r+\binom{a_k-1}{r-1}\\ &\ge\binom{a_i}r+\binom{a_k-1}r+\binom{a_i}{r-1}...


2

Since $g$ and $h$ are continuous, $g^{-1}(U)\cap h^{-1}(V)$ is an open set containing $x$ and hence it contains an element $a$ of $A$. But $a\in g^{-1}(U)\cap h^{-1}(V)$ implies that $f(a)=g(a)\in U$ and $f(a)=h(a)\in V$ which contradicts the assumption that $U,V$ are disjoint


2

for $|x|<1$ : $$ \frac{1}{1-x}=\sum_{n\geq 0} x^n $$ put $t=-x$ you get : $$ f(t)=\frac{1}{1+t}=\sum_{n\geq 0} (-1)^n t^n=\sum_{n\geq 0} \frac{(-1)^nn!}{n!} t^n=\sum_{n\geq 0} \frac{f^{(n)}(0)}{n!} t^n $$ so $$ f^{(n)}(0)=(-1)^n n! $$


1

I think you have your wires crossed; particularly it seems like you're thinking about how the sum of the first $n$ odd numbers gives $n^2$. This is woefully incorrect. Here is how you should frame it: Suppose that $x_1,\ldots, x_n\ge 2$, then you want to show that $x_1\cdots x_n$ is odd if and only if $x_i$ is odd for all $i$. In terms of an induction ...


1

the result is correct but in the demonstration it is best not to confuse the symbol $\varepsilon$ and $\delta$ Since $f$ is uniformly continuous then $\forall\epsilon>0, \exists\delta>0$ s.t. $\forall x,x'\in X, d(x,x')<\delta\Rightarrow\rho(f(x),f(x'))<\epsilon$ But since $(x_n)$ is Cauchy in $X$, then $\forall\delta>0, \exists N$ s.t. $\...


1

The proof of your claim comes down to proving whether $\pi_X\left ( \bigcup_iW_i\right ) = \bigcup_i \pi_X(W_i)$ where the $W_i$ are product open sets. Clearly if $U\times V$ is a product open set in $X\times Y$ then $\pi_X(U\times V) = U$. $(\Longrightarrow)$ Let $p \in \pi_X\left ( \bigcup_i U_i \times V_i\right )$, then there exists an element $q \in ...


1

I think it might raise eyebrows for those who are new to set theory and topology, but that is precisely why I like this notation. One better get used to certain sets having elements which are sets containing further elements themselves. That said, if you want to “translate” Let $U$ be an open set that contains $x$ a better solution in this case ...


1

much better to prove the contrapositive $a$ odd implies $a^n$ odd. This is easily proved by induction since the product of two odd numbers is odd.


1

Suppose $$A=B+C$$ If $$B^T=B, $$ $$C^T=-C,$$ then according to the known property of transposition of sum of matrices $$A^T=(B+C)^T=B^T+C^T=B+(-C)=B-C$$ Now we have $$A=B+C \tag 1\\ $$ $$A^T=B-C\tag 2\\$$ Adding $(1)$ to $(2)$ gives $$B={(A+A^T)\over 2}\\ $$ Subtracting $(2)$ from $(1)$ gives $$C={(A-A^T)\over 2}\\ $$



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