$a_n=\frac{n+1}{n}$ has a ratio of $\frac{n(n+2)}{(n+1)^2}=\frac{n^2+2n}{n^2+2n+1}$ and converges to $1$, if I understand what you're looking for correctly.
Dear mathematicians, can you please help me fine tune the symbolic translation of UDHR Article 1? "All human beings are born free and equal in dignity and rights. They are endowed with reason and conscience and should act towards one another in a spirit of brotherhood." My draft version:
"An open subset $R$ of $X$ is regular open if $(\overline{R})^\circ = R$. The collection of all regular open sets is denoted $RO(X)$. It is well known that $RO(X)$ is a complete Boolean algebra under suitable definitions of sup and inf." What are those suitable definitions?
Given a homeomorphism $f:(0,1)\rightarrow\mathbb{R}$, is it possible to construct $f$ such that $f(R)$ covers countably infinitely many natural numbers, where $R$ is an arbitrary subset of $(0,1)$? My instincts tell me that no, such a construction isn't possible, because $\mathbb{N}$ is not dense in $\mathbb{R}$, but I'm kind of stumbling on justifying it to myself.
@Rithaniel So get your quantifiers straight. Of course, given $f$, there are lots of choices of $R$ that work. But you seem to want to be given $R$ and find an $f$. (So this is badly stated.) If $R$ is a finite set, it can't work. So what is your question, really?
Okay, the idea is that we construct $f$ to be a homeomorphism with the property that any (let's say uncountably infinite) $R$ that we can possibly select will necessarily have an image under $f$ which contains countably infinitely many natural numbers.
Slightly. So you start with an arbitrary uncountable set $R$. Actually, I think all you need is a countable (infinite) set $R$. Then you wish to construct a homeomorphism $f$ so that $f(R)$ contains infinitely many natural numbers.
suppose f,g are differentiable on (a,b) and suppose g'(x)=f'(x) for all x in (a,b). How do I prove that there is a c in the reals such that g(x)=f(x)+c
Yeah, C'. I think I have finally understood why the C'(1/6) condition is so important: in the van Kampen diagram (at least I think that's what they are called? Lyndon & Schupp just calls them "maps") of such a group, the regions will have at most 1/6-th of overlap compared to the length of the boundary of either regions. So the tightest such thing you can imagine is a "hexagonalation" of an open subset of the plane, where any two hexagonal regions share a side, exactly 1/6-th of the boundary.
Pass to the dual graph, which is a triangulation, for ease. If you take the boundary curve of the triangulated region, the total curvature around that is exactly $2\pi$.
At each vertex $v$ of the boundary curve, the interior angle is $(d(v) - 1) \cdot 2\pi/6$ where $d(v)$ is the degree of $v$
The curvature at $v$ is $\pi$ minus that, which is $(4 - d(v)) \cdot 2\pi/6$.
Sum of that over all boundary vertices is the total curvature, which leads us to $\sum_{v \in \partial D} (4 - d(v)) = 6$
But since this is the tightest such scenario, for the van Kampen diagram of a C'(1/6) group this is going to be $\sum_{v \in \partial D} (4 - d(v)) \geq 6$
Intuitively speaking, at least. They arrive at this formula through much manipulation, but this is clearly some kind of curvature restriction on the group
And in retrospect this becomes obvious as in the dumbest case you have the hexagonal tiling of the plane; if degree of each face is more than 6 ('cuz in C'(1/6) you demand the pieces have length proportion < 1/6 not <= 1/6), the mental picture is that of regular $n$-gons tiling $\Bbb H^2$, for $n > 6$
Let's say I have two commutative rings $A, B$ and a ring epimorphism $\phi : A \to B$. this induces a continuous map $\phi^* : \operatorname{Spec}(B) \to \operatorname{Spec}(A)$. Choose $x \in \phi^{-1}(f)$, how can I show that for $\mathfrak{q} \in \overline{V(\{x\})}$ there exists a $\mathfrak{p} \in \overline{V(\{f\})}$ such that $\mathfrak{q} = \phi^{-1}(\mathfrak{p})$?
That line above is meant to denote complement not closure
