# Tag Info

## Hot answers tagged definition

5

If you take the usual definition of continuity and negate it, you get $$\exists \epsilon > 0 \forall \delta > 0 \exists c\in \mathbb{R}: |x - c| < \delta \text{ and } |f(x) - f(c)| > \epsilon.$$ Your definition is missing quantification on $c$, but more importantly, you've switched the orders of the quantifiers on $\epsilon$ and $\delta$, ...

4

You can "weaken" it that way (it turns out to not be a weakening at all, except in apparence). Even take $N=1$ (or any fixed positive number) if you want. Clearly, the first definition implies the second. Now, for the converse: assume we have $\exists N>0 \text{ s.t. }\forall \varepsilon \in (0, N), \exists \delta > 0 s.t. |f(x) - L| < \varepsilon ... 3 As stated it doesn't appear to work. For simplicity set$S = \mathrm{Spec} (k)$. Here's a counterexample: let$X$be any$k$-variety,$e : S \to X$any point of$X$, and$\rho : X \to X$any morphism. Set$\mu : X \times X \to X$to be the constant morphism$\mu(x,y) = e$. This clearly satisfies properties (1) and (2) that you listed (since ... 2 Hint: Assume$(x_n)$is C-convergent to$x$. Fix$n > N$. Then for all$\epsilon > 0$we know that$0 \leq d(x_n,x) < \epsilon$. So$d(x_n,x) \in \cap_{\epsilon > 0} [0,\epsilon) = \{0\}$, which is to say that$x_n = x$. Thus a sequence$(x_n)$is C-convergent if and only if there exists$x \in \mathbb{R}$and$N$such that$x_n = x$for all$n ...

2

The definitions are equivalent. Assume that there exists $N > 0$ such that for any $\varepsilon \in (0, N)$ there exists $\delta > 0$ for which $0 < |x - a| < \delta$ implies that $|f(x) - L| < \varepsilon$. We want to show that given $\varepsilon' > 0$ there exists $\delta' > 0$ for which $0 < |x - a| < \delta'$ implies that ...

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Your intuition is pretty good. You actually claimed that it doesn't matter what the function does "far away" from $f(x_0)$ (More precisely, outside some neighborhood of $f(x_0)$).

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$C$-convergence is much stronger than convergence. The sentence for every $\epsilon>0$ and every $n>N$, $d(x_n,x)<\epsilon$ is the same as for every $n>N$, $x_n=x$.

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I believe Frege's The Foundations of Arithmetic is considered to be a classic in the subject.

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There are several ways to construct a fully functional axiomatic theory of mathematics. I would recommend you to start your journey with Set Theory, which has been the classical framework for math since early in the past century. 'Naive Set Theory', by Halmos, is a short but very complete introduction to the topic. You may also want to dabble a bit in ...

1

Your question betrays muddleheadedness, but that is fine, because the cure of this disease is a most likely result of your quest. Believe it or not, there are people who don't think the foundation of mathematics lies in logic. The formalist school believes that mathematics is simply ink marks on paper; and I think they are the majority as of today. If you ...

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Since $U=sp\{u_1,u_2\}$ is a plane, then any span where $sp\{u_1,v_1\}: v_1\perp u_1$ doesn't necessarily span the same plane as U. So we we can't just take any vector that's orthogonal to $u_1$ as $v_1$. Thanks Gerry Myerson.

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Yes. Suppose $\lim_{x\to+\infty}f(x)=L$. That is, for every $\epsilon>0$ there exists $x_0$ such that $x>x_0$ implies $|f(x)-L|<\epsilon$. Let $x_n\to+\infty$, that is for every $M>0$ exists $n_0\in\Bbb N$ such that $n>n_0$ implies $x_n>M$. I claim that $f(x_n)\to L$. Indeed, let $\epsilon>0$ be given. Then pick $x_0$ such that ...

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Only for almost all $x$. This follows from the definition of the "norm". One possible (equivalent) definition is that $\|f\|_\infty$ is the smallest number $M>0$ such that $|f(x)|\le M$ for almost all $x$.

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The "functions" in $L^p$ spaces, including $L^\infty$, are actually equivalence classes, where $f$ and $g$ are equivalent if they are equal almost everywhere. So any statement you make about $f(x)$ can only be interpreted in the "almost everywhere" sense. However, for $f \in L^\infty$ you can choose a representative of the equivalence class that is bounded ...

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