Can anybody provide me with an example of a nonempty set $G$ which is closed under an associative product $\cdot: G\times G\to G$ with right identity ($a\cdot e = a$) and left inverses (There exists $y(a)$ such that $y(a)\cdot a = e$) which is not a group?
Ah you're right. My intuition ignores the gray area that a set can be both closed and open (containing all of its limit points, but also such that every point has an $\epsilon$-ball contained in the set)
@Jakobian This is what is confusing to me. This makes me think that a set is closed iff it contains all of its limit points, so $\bar{E} = E$. But I thought there was something wrong with that intuition
Looks like (at least) my second statement is false; the set $E = \{1/n \mid n\in\Bbb N\}\subset \Bbb R^1$ has the single limit point $0$ which isn't contained in $E$, yet the set is closed.
@onepotatotwopotato wow, so my misunderstanding runs deep. Are there any necessary and sufficient conditions for openness and closedness involving the containment of limit points?
Very basic point-set question: I got the (unconfirmed) intuition that a convenient distinction between closed and open sets could be that a set $E$ is closed iff it contains all of its limit points. Conversely, $E$ is open iff there are limit points which are not contained in $E$. Is this a valid understanding? Like, could I use these facts in a proof?
@CiurkitboyN Yep! If an event $E$ has a probability of occurring $0 < P(E) \leq 1$ in some time unit $t$, then we can expect $E$ to occur about once every $t / P(E)$ time units.
In "open cover," the adjective "open" is modifying the sets which compose the cover, while in "finite subcover," the adjective "finite" is modifying the size of the family of sets which compose the cover.
@leslietownes understood. My main misunderstanding was the phrasing, b/c I thought that if any finite open cover of a set $K$ existed, then it is compact (which turns out to be a very weak criterion, because basically every set has a finite open cover)
Based on this understanding (which I only gather from looking at the definition of compactness long enough), there are no non-compact spaces at all, so clearly I must be missing something... can anybody point out what I'm getting wrong?
If there's some rule that we can't use a space $X$ as its own finite open cover, then we could just split $X$ in half with overlap (i.e. $\Bbb R$ could use $(-\infty, 1)\cup(-1,\infty)$) and still find a finite open cover.
Can anybody explain why $\Bbb R$ is not compact by definition, but the (closed) unit disc is compact by definition? When I see the construction that $\Bbb R$ is not compact because we can find an infinite open cover with no finite subcover, I am confused because one can do the same thing via the proper choice of pathological open cover for the closed unit disc. Likewise, one can find a finite open cover of $\Bbb R$ ($\Bbb R$ itself is open, so we can just use itself as a finite open cover).
Can anybody testify to the truth of this statement? "If $R$ is a commutative ring, then $R+r := \{r + x \mid x\in R\}$ is isomorphic to $R$." It feels true to me when I think about it, but IDK if it's proven
@leslietownes Yeah this was the main thing, I posted it and them almost immediately got a lot of downvotes, it freaked me out. I guess you're right I didn't prove it satisfied the homomorphism criterion
Just for solution verification purposes, can anybody vouch for the faithfulness/unfaithfulness of the fraction field functor? ($\text{K}:\mathbf{Int}\to\mathbf{Field}$)
Is this counterexample (pair of non-parallel morphisms in $\mathbf{Int}$ mapping to the same morphism in $\mathbf{Field}$) enough to prove unfaithfulness of the fraction field functor $\text{K}$?
This scaling homomorphism $\Lambda$ has the same image morphism in $\mathbf{Field}$ as the identity morphism $\text{id}_R : R\to R$, because they both map to the identity morphism $\text{id}_K$ in $\mathbf{Field}$ where $\text{K}:R\mapsto K$
I'm specifically goofing around with the fraction field functor $K:\mathbf{Int}\to\mathbf{Field}$ which maps the category of integral domains with injective homomorphisms between them to the category of fields with injective homomorphisms between them. My claim is that this functor is not faithful, because for every integral domain $R\in\mathbf{Int}$, there exists a "scaling" homomorphism $\Lambda:R\to R'$ via $r\mapsto \lambda r$ for non-zero, non-unit $\lambda\in R$.
In proving that a functor $F:A\to B$ is not faithful, does it suffice to find a single pair of morphisms between two objects $f:A_1 \to A_2$ and $f':A_3 \to A_4$ which the functor maps to the same morphism in $Ff=Ff'=g: B_1 \to B_2$ in $B$?
Does anyone with an intuition for Abstract Algebra have a useful way of remembering the distinctions between maximal ideals, principal ideals and prime ideals of rings? When I was first learning about ideals, I thought they could just be considered the ring-equivalent of "normal subgroups" in group theory, but with the increased complexity of rings > groups, there are these other noteworthy varieties of ideals in the form of prime/principal/maximal ideals, and I often mix them up
If an $n\times n$ matrix, $\mathbf{A_n}$, is populated with integers randomly distributed on {-1, 0, 1}, what is the resulting probabilistic distribution of the determinant $\det(\mathbf{A_n})$? Essentially, if $X ~ \det(\mathbf{A_n})$ as described above, what is the pdf of $X$?
