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The Principle of Mathematical Induction says that for all "properties" $P$, $$\left(P(0)\land\forall k\in \mathbb N\left(P(k) \implies P(k+1)\right)\right)\implies \forall n\in \mathbb N(P(n)).$$ So you're basically asking how to write the $\forall k\in \mathbb N\left(P(k)\implies P(k+1)\right)$ bit. It's a universal statement. It's common to start those ...

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Given an injection $f : S \to \mathbb{N}$, where $S$ is infinite, you can define a function $h : \mathbb{N} \to S$ by, for each $n \in \mathbb{N}$, defining $h(n)$ to be the element $s \in S$ for which $f(s)$ is the $n^{\text{th}}$ least element of $f(S) \subseteq \mathbb{N}$. This is possible because $S$ is infinite, so such an element $s \in S$ always ...

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Consider the problem systematically: Starting with $1$, the sequence cannot contain the number $2$ until there are $2$ adjacent $1$s. Similarly, the sequence cannot contain $3$ until there are either $3$ adjacent $1$s or $3$ adjacent $2$s. Since the former will occur first, this occurs when there are $3$ adjacent $1$s. Continuing this trend, it is not ...

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Welcome to the wormhole. A set which has a countably infinite subset is called Dedekind Infinite. It is not provable within Zermelo-Fraenkel set theory (ZF) that every infinite set is Dedekind finite. This was demonstrated by Paul Cohen, who exhibited a model of ZF in which there is an infinite, Dedekind-finite set of real numbers. See Asaf's answer to ...

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Consider the range of the injection $f:S\to\Bbb N.$ It will necessarily be an infinite (why?) subset of $\Bbb N,$ which can readily (and order-isomorphically!) be mapped onto $\Bbb N$ by sending its least element to the least element of $\Bbb N,$ its second-least element to the second-least element of $\Bbb N,$ and so on. Edit: Basically, all you have to do ...

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Since $f$ is an injection, we have that $f:S\rightarrow f(S)$ is a bijection. Since $f(S)\subset N$. It suffices to find a bijection between some infinite subset $A\subset N$ and $N$. Now, we can construct the following bijection $g:N \rightarrow A$: $g(1):=\min A$, $g(2):=\min A-\{g(1)\}$, $g(3):=\min A-\{g(1),g(2)\}$ and so on. It remains to show that ...

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You're proof seems fine. You could be a little more succinct. For example, Since, $\sum |c_n|$ is a sequence of non-negative terms, it suffices to show that $\{C_n\}$ is bounded. \begin{align} C_n & =\sum_{k=0}^n\left|\sum_{m=0}^k a_mb_{k-m}\right|\\ & \le\sum_{k=0}^n\sum_{m=0}^k|a_mb_{k-m}|\\ & =\sum_{m=0}^n|a_n|\sum_{k=0}^{n-m}|b_k|\\ ... 1 The deeper we dive into the ocean of mathematics, the stronger is my conviction that we actually do not know anything. Let me give some examples: 1. Every polynomial p(z) has a root in \mathbb{C} 2. For a Hilbert space H, H=W\oplus W^{\perp}, where W is a closed subspace of H and W^{\perp} is its orthogonal complement. 3. Every subgroup ... 1 You can choose \varepsilon as you please, then you determine an N such that &c. Completion of the proof:\lvert s_n-s\rvert<\varepsilon\iff s-\varepsilon<s_n<s+\varepsilon.$$In particuler, \; s_n>s-\varepsilon=s-\dfrac{s-a}2=\dfrac{s+a}2>\dfrac{2a}2=a. 1 Do you know that for a chosen \epsilon>0 there is N s.t. for all n\ge N the terms of the sequence s_n belong to the interval (s-\epsilon,s+\epsilon)? Now to answer the question it suffices to choose \epsilon such that s-\epsilon>a. Draw a picture! 1$$(X \cap Y) \setminus (X \cap Z) = (X \cap Y) \cap (X \cap Z)^C = (X \cap Y) \cap (X^C \cup Z^C) \\ = (X \cap Y \cap X^C) \cup (X \cap Y \cap Z^C) = X \cap(Y \cap Z^C) = X \cap (Y \setminus Z) $$1 It’s generally easiest to start with the more complicated expression, which in this case is the indicator function corresponding to the righthand side,$$(1_A+1_B-1_A\cdot 1_B)\cdot(1_A+1_C-1_A\cdot 1_C)\;.$$If you multiply this out, you get$$\begin{align*} 1_A\cdot(1_A&+1_C-1_A\cdot 1_C)+1_B\cdot(1_A+1_C-1_A\cdot 1_C)-1_A\cdot ...

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Hint You must supplement your textbook with some book on set theory, like e.g. : Patrick Suppes, Axiomatic set theory (1960 - Dover reprint). See page 22 for the usual definition of the "inclusion" relation : $A \subseteq B \leftrightarrow \forall x ( x \in A \to x \in B)$ and page 47 for the definition of the power set : $\mathcal P(A) = \{ B : B ... 1 If by infinite, you mean Dedekind-infinite, then there exists an injective function$f:S\to S$such that$f$is not surjective. Since$f$is not surjective, there exists$a_0\in S$such that, for all$x\in S$we have$f(x)\neq a_0$. Then the set$\{a_0, f(a_0), f(f(a_0)) \cdots \}$can be shown to be order-isomorphic to$N$. No need for AC. 1 For #1, that doesn't work, because you don't know that$d_B$is induced by a norm. In fact it is not, because all norms on positive dimensional spaces are unbounded, but that metric is bounded. (I might edit if you post an attempt and show where you get stuck.) For #2, one open cover that you always have of a set$A$in a metric space is$$\{ B_\epsilon(x) ... 1 As noted in the comments, the problem has been misread. Here's an (inelegant) solution the problem. This is one of those sad coordinate-geometry arguments that illuminates very little about the geometry going on here, but can be done with bullet-headed algebraic computation. If the statement of the problem is true, the constant value must be: ... 1 This is a typical proof by induction. The base case is just that the first term ($1$) does not contain any digit greater than$3$nor more than$3$consecutive, equal digits. If some term had a digit greater than$3$, it would have had to come from more than$3$(that is,$4\$ or more) consecutive, equal digits in the previous term. But the first four of ...

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