Try with an example. If, for istance, we have: $y=\frac{x^2-3}{x-1}$ To do what I need, we need to put: $\frac{x^2-3}{x-1} > 0$ ||$x^2-3>0$ ||$x-1 > 0$ (the two vertical lines are the symbol for the "false system")
@unNaturhal I’d simply chop the real line into intervals at the points where the numerator or the denominator are $0$: $-\sqrt3,1,\sqrt3$. Then I’d check the sign of each factor on each interval and combine them. I’ve not seen the term false system before.
Exactly. Putting the solutions of the "false system" on the real number line, you'd have something like this: $-\sqrt{3}$ $1$ $\sqrt{3}$ _______|________|________|________ +++++++--------------------------++++++++ ------------------------++++++++++++++++ - + - +
So, you just take the areas where there are "+" ('coz you have set up the function">" than 0)
I think that I’d get a bit excited $-$ frenzied, even $-$ if someone tried to shove me into an oven, but I’ll admit that if they succeeded, the excitement would be shortlived.
I guessed one or the other, since $\beth_\omega$ could be formed by taking the union of an appropriate set... hm. I'll have to think about this a bit more. Thanks!
That reminds me: I was going to give you (@Asaf) my favorite non-algebraic example of an inverse limit. Let $\mathscr{U}$ be a free filter on $\omega$. For $U\in\mathscr{U}$ let $X_U$ be a copy of $\omega+1$ with the following topology: each $n\in\omega$ is isolated, and basic open nbhds of $\omega$ are sets of the form $\{\omega\}\cup(U\setminus F)$, where $F$ is finite. $\langle\mathscr{U},\supseteq\rangle$ is a directed set.
For $U,V\in\mathscr{U}$ with $U\supseteq V$ let $\pi_{UV}:X_V\to X_U$ be the identity map; note that $\pi_{UV}$ is continuous. Let $X$ be the inverse limit of this inverse system; then $X$ is homeomorphic to $\omega+1$ with the topology in which each $n\in\omega$ is isolated, and $\mathscr{U}$ is the family of open sets containing the point $\omega$.
@Matt: Tips to improve the answer: explain that Q[i] is a field because inverses can be exhibited $$(a+bi)(a-bi)=a^2+b^2\implies(a+bi)^{-1}=\frac{a}{a^2+b^2}+\frac{b}{a^2+b^2}i.$$
Second, Z[i/2] is not a field because the elements are of the form $$u_0+u_1\frac{i}{2}+u_1\frac{i^2}{2^2}+\cdots+u_m\frac{i^m}{2^m}=\frac{2^mu_0+2^{m-1}u_1i+\cdots+2u_{m-1}i^{m-1}+u_mi^m}{2^m}.$$ so only have power-of-2 denominators, so e.g. 3 has no inverse.
You say "you can't write 1/3 in terms of elements from Q." I don't follow; can't you just write 1/3 because it is an element from Q? Anyway, my understanding of the OP is that the question was "why isn't this a field like Q[i] is?"
Thought so. Also, even though Q[i] and Q(i) turn out to be the same thing, the OP is referring to Q[i]. The latter is a field by definition, the former requires demonstration of inverses at least.
@anon So $Q[i]$ (read "Q adjoin i") is the set of all $a + ib$ by definition. And $Q(i)$ is the field of quotients or how do I have look upon this difference?
@BrianMScott Exactly. : )
@anon Well, I had a different definition in mind. I thought that $[\cdot]$ is used when it's just a ring. And $(\cdot)$ when it's a field.
I should be doing set theory and here I am, answering questions on SE.
@Matt: R[x] refers to the finite polynomials a+bx+cx^2+... This works when R is a ring, and fields are special cases of rings so R can also be a field. With i^2=-1, all polynomials in $i$ reduce to the form $a+bi$. Now for K a field, K(x) refers to the smallest field containing K and x. We have Q[i]=Q(i) because inverses of a+bi can also be written in this form. Something like $\mathbb{Q}[\pi]$ is not a field however, even though $\mathbb{Q}(\pi)$ is.
$K(x)$ can be constructed as the field of fractions of $K[x]$ of course
When $x$ is algebraic we can speak of K(x) and K[x] interchangeable though. Since the norm defined by multiplying automorphisms of an element evaluates to elements in the base field, we can divide the product out by the element we want to invert and exhibit an inverse, same as we do for a+bi in Q[i].
@mk It does. But it’s more a difference in orthographic custom than a real difference in the way compounds are constructed: German simply runs them together in many cases when English doesn’t.
@anon I once heard that in English, if you spell the first few letters and the last few letters correctly people are still able to figure out the word. for example: beutiful
The notion of word isn’t actually well-defined cross-linguistically. There really doesn’t seem to be any definition that works really well for all languages.
@BrianMScott I heard that English is among the most difficult languages to learn to use properly and I guessed it was partially because of the large number of idioms.
Whether a language is difficult to learn depends enormously on what language(s) you already know. It’s not clear that there is an absolute scale of difficulty. Different languages have different difficult areas for different speakers.
Depends on the kind of history; social history, military history, diplomatic history $-$ they have much in common, but they also call on different specific background knowledge. (And that was by no means an exhaustive list.)
