# Tag Info

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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