Ya I still don't quite see what the issue is Thorgott, but likely my error. I've called the sets we're quantifying over card $A$ and card $B$, but they're just sets we're quantifying over too?
it's an abuse of notation, which is generally something completely fair and reasonable to do, but if you're going out of your way to do propositional logic, it's not
in any case, even if you fix this syntactical error, it won't fix the semantic error
cause there is only supposed to be one fixed choice of card A for every set A
Maybe I should say the statement that (it is true that) $$\forall A [\exists C [(A \approx n \in \omega \implies C = n) \land (\forall B [\exists D[A \approx B \iff C = D]])]]$$
@EE18 I think that part of the problem is that Enderton very specifically did not try to give you a statement which is meant to be formalized in this way. The "Promise" he gives is informal, and is meant to give you some intuition to hang your hat on until he can get to the formal definition.
An “ordered extension field”…does that basically mean an ordered field for which there is an embedding of the first ordered field into the “bigger one”. My book speaks about subsets but want to clarify
Am on mobile right now so lord forgive me for my sins, here goes:
We have seen that the equation x2 = a for positive a is, in general, not solvable in Q. Since x2 is the area of a square with side x, this means, for example, that there is no square of area 2 — so long as we stay within the field of rational numbers. As is known from high school, in order to remedy this unsatisfactory situation, we must allow squares with sides whose lengths are ‘irrational numbers’.
This means that our field Q is too small, and we need a larger field which contains Q as a subfield, and in which the equation x2 = a for a > 0 always has a solution. In other words, we seek an ordered extension field of Q in which the equation x2 = a is solvable for each a > 0.
Sorry that is copy pasted from my phone
The book is very imprecise, this is the first place “ordered extension field” is used
I’ve only given the definition above because of what I’ve seen in enderton
yeah, they mean an embedding of Q in an ordered field (if they haven't defined this, maybe they will later; it's not a field with a notion of > that interacts with the field operations in expected ways)
some folks on main are particularly sensitive about the use of that tag, and there is sometimes a close question as to whether 'is there a way of doing this better' is even a question about math within the scope of the site.
i don't think people have any deep seated objection to proof verification as a concept (which is why the tag exists), but, use of that tag often correlates with very low effort questions. some users feel that you can drape the tag and "is this right?" over almost anything and make it compliant with the site, and... no.
I am looking at the quesiton a bit more closely and it seems clear that it will be an example of "depends what foundations you're working with" and so maybe i won't waste folks time here. will save the goodwill for the numerous other things i ask about...
for me personally, the thing is not so much "is it a proof verification or not," but "has the asker actually thought about the question" and "is the question answerable without digesting a mountain of background"
maybe with some thought it will get shorter. :) it is sometimes possible to wrap details in little packages. if you can turn parts of the argument into lemmas that you are willing to take for granted, maybe the details don't matter.
proof verification is no fun, i am sure it is very helpful for the OP, but any help is very case specific so there is little leverage. it is more like one on one consultation.
@onepotatotwopotato Actually that is probably correct. Smooth in the real sense, yes -every smooth manifold is a nonsingular algebraic variety. There are complex manifolds which are not algebraic however
@JohnZimmerman why every smooth real manifold is a nonsingular algebraic variety?
I actually don't know what algebraic variety is. I'm just curious because I've heard someone saying before that "...smooth manifolds, more generally algebraic variety..."
Consider a group $G$, (picture below). We have $8$ vertices. We have $96$ white strands. Consider the following group actions a) rigid rotations (on vertices) and b) modular rotations (on white strands) and c) rubiks cube rotations (rotating along the "rubber band curves") Rigid rotations form the octahedral group. Modular rotations are defined as rotations of the white strands stabilized by exactly one vertex. We can compose these $3$ transformations together.
maybe I am misunderstanding but it seems to me that the elements are of different dimension
@Jakobian This is, I think, the more salient point. The example that is more solid in my mind is that any group can be realized as the fundamental group of some topological space (by cleverly gluing together $n$-cells), but any particular group may be the fundamental group of many different non-homeomorphic spaces.
@JohnZimmerman I thought you were answering my question?
I suppose that it is possible that every group can be seen as acting on some geometric space (I'm not quite sure what this means---it is, as @Jakobian says, quite vague), but I am not sure what we get from that.
Consider the $\sigma$-algebra of countable or co-countable sets, i.e. the collection of subsets of $X$ that are countable or whose complement is countable. Does the set $X$ have to be uncountable? This is from Folland's book.
@psie If $X$ is finite, then the collection of countable or co-countable sets is just the powerset. So no, $X$ does not have to be finite, but, like, it is boring in that case.
@psie I wasn't commenting on the relationship between them
When you mix general topology and measure theory, that's very interesting in itself
But I was commenting on the fact that considering a set with a family of certain subsets is very similar in spirit
But with measure theory you are, from the very beginning (and this is what motivates the subject) invited to consider set theory and foundations and how problematic is it when trying to study analysis
This is also seen in definition of sigma-algebra
In general topology, the subject is much simpler in this regard
@psie if you say "the $\sigma$-algebra generated by the open sets" I agree, but if you do the transfinite recursion construction to get the stratified hierarchy, then I'm not sure I agree with its unconstructiveness
@Jakobian I'm not sure that I know of a term off the top of my head. Generally, there is not much point in talking about such an object---either the topology is compatible with the measure (so, like, you have Borel measure on a topological space), or the topology is not compatible with the measure, and bad things happen.
Consider the vector space $M_2(\Bbb R)$ and consider the following subset: $$V_n = \left\{ \begin{pmatrix} a & b \\ c & 0 \end{pmatrix} \in M_2(\mathbb{R}) : b = -na + c \right\}, \quad \text{with } n \in \mathbb{R}$$$(a)$ Verify that $V_n$ is a subspace of $M_2(\Bbb R)$ for every $n \in \Bbb R.$
I need to show that it satisfies the three properties that define a vector subspace: 1) $V_n$ contains the null vector. 2) $V_n$ is closed under vector sum. 3) $V_n$ is closed under multiplication by a scalar.
I have a doubt about what I did next, now I'm writing
If I wanted to determine the size $V_n, \forall n \in \Bbb R$
The dimension of a vector space is effectively the number of linearly independent vectors that generate it.
@Pizza I would encourage you not to use multi-line comments in chat. When they get long, they gain a scrollbar, and it is not possible to see the entire comment all at once. It is frustrating to read. I would suggest breaking things up. E.g. ask the question, then give your responses to each of the three parts in separate comments.
@Pizza I see no reason to delete it. I am just expressing the fact that what you a have posted doesn't fit, making it harder to read. Personally, this makes me less inclined to bother to read it.
It seems to me that you have concluded that $V_n$ is a vector subspace of $\mathbb{R}^2$ for any $n\in\mathbb{R}$. That was what the exercise asked you to show. So you are done, n'est-ce pas?
If I wanted to determine the size $V_n, \forall n \in \Bbb R$ The dimension of a vector space is effectively the number of linearly independent vectors that generate it.
If i sub my constraint for b, I would get a matrix depending on $a$ and $c$. And then i can split it into a linear combination of $a$ and $c$ and determine a basis.
@AlessandroCodenotti I apologize, I was not trying to argue with you. I was just adding another data point. The phrase "Let $(X,d,\mu)$ be a metric measure space" shows up a lot in my thesis.
@Pizza So you don't really care about any of the exercise you reproduced above, you are just asking about the dimension of that space, yes?
If i sub my constraint for b, I would get a matrix depending on a and c. And then i can split it into a linear combination of a and c and determine a basis.
@AlessandroCodenotti yes but I was looking for a word for topological spaces
but while I have you here
I remember there was a word for spaces $X$ such that $d(X) = |X|$ i.e. the density agrees with cardinality
do you remember how they were called?
or maybe I'm wrong about there being such word
what the person was asking is a particular case of, does there exists a topological space $X$ with $c(X) = \omega$ and $d(X) = |X| = \kappa$ for every cardinal $\kappa$
I believe this is false
Certainly its non-metrizable for $\kappa > \omega$ and if such space is Hausdorff then $|X|\leq 2^{\chi(X)}$ which means that the character of $X$ needs to be large if $\kappa$ is large
this would also means that the weight $\chi(X)\leq w(X)$ would have to be relatively large
Having chosen a value $n \neq 0$ , determine a basis $B$ of $V_n$. Theoretically, the one I obtained before would already be fine as a basis for $V_n$, right? this:
