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 Revival
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1d
comment A subgroup of special linear group
I've never seen a name for this group (not have I seen this group explicitly discussed before, even without a name).
Jul
4
comment Proof that intervals of the form $[x, x+1)$ or $(x, x+1]$ must contain a integer.
Reformulation of @MattSamuel's comment: How would you even define the fractional-part function if you didn't already know what the OP is trying to prove?
Jul
2
awarded  Revival
Jul
2
comment Well ordering of type epsilon one
@AsafKaragila I don't think this characterization of $\epsilon_0$ is right. The ordering of type $\epsilon_1$ that I describe in my answer looks primitive recursive. In fact, I think all the recursive ordinals occur as order-types of primitive recursive wellorderings. More generally, you can vary the complexity of the well-orderings all the way from primitive recursive up to hyperarithmetical without changing the resulting order-types.
Jul
2
answered Well ordering of type epsilon one
Jul
2
comment I don't understand what my Calculus hw question is asking of me…not looking for answers, just guidence.
The opp/hyp definition that you mentioned makes sense only for angles that can occur in a right triangle, not for obtuse angles like $3\pi/4$. Look in your book for information about trigonometric functions for angles that are not in the range from $0$ to $\pi/2$. You'll probably find something very similar to @pjs36's comment.
Jul
2
comment First Order Logic Question
Since you began with "if by tautology you mean ...," let me fill in that, on the other hand, if by "tautology" you mean "valid for purely propositional reasons" (which is what I would ordinarily mean), then the answer is "no". (I expect though, that what you wrote is what the OP meant.)
Jul
2
answered Is defining an “$S$-scheme” this way just a stylistic choice?
Jul
1
answered Why are conformal mappings necessarily 1 to 1?
Jun
30
comment Is this operator $A = \pmatrix{1&1\\0&1}$ self-adjoint?
Another way to see that no scalar product can make $A$ self-adjoint is to notice that both of its eigenvalues are $1$ yet $A$ is not the identity matrix.
Jun
30
comment Verify that $A \oplus B$, where $A$ and $B$ are cyclic groups of orders 2 and 3, is the cyclic group of order 6
In fact, once you realize that an element of $A$ can't work, because the subgroup it generate would be included in $A$, and an element of $B$ can't work for an analogous reason, then the only elements left to check are the two that work --- so you'd have "trial without error."
Jun
30
comment Verify that $A \oplus B$, where $A$ and $B$ are cyclic groups of orders 2 and 3, is the cyclic group of order 6
Finding a generator is pretty easy by trial and error. There are only 6 elements to check, and 2 of them work. (And $b$ does not work.)
Jun
26
comment Prove the connected components are uncountably many
This shows not only that the graph is connected but also the stronger result that its diameter is $2$.
Jun
25
comment Teaching to Learn
I've found that teaching is an excellent way to learn mathematics, but if you take "teaching" literally, then you need someone to teach. Even though I'm a professor, I can't always find someone to listen to the particular things that I want to explain. My solution to this difficulty is to give lectures without an audience --- for example, in the shower, or lying in bed. You can even give yourself lectures while just walking down the street if you don't mind other people thinking you're weird. The lectures can even be accompanied with vigorous hand-waving (looking even weirder).
Jun
25
comment Is the game fair or unfair?
A more explicit response to Noah's second comment: If the first ball drawn is red but the next two are black, then, according to the rules you gave in the question, the player wins. There is a time, namely after the first three balls are drawn, when more black than red balls have been drawn, namely two black and one red.
Jun
25
comment Homotopy equivalence of $S^{2} \vee S^1$ to $S^{2} \cup A$ where A is a line segment joining noth and south poles
I don't understand "the intersection of these two objects is path connected." If the two objects are $S^2$ and the line mentioned in the immediately preceding sentence, then their intersection consists of two points, so it is not path connected. If the two objects are something else, then what are they? (Also, you seem to be trying to prove that the two spaces in the problem are not homotopy equivalent, but in fact, as the problem says, they are.)
Jun
22
comment May two or more bases of different sizes generate a same topology?
@Lin You probably meant $B1'$ when you wrote $B1$, as there is no $B1$ in your question. The lemma works with $B$ and $B1'$: For each point $x$ in your space (i.e., $a$ or $b$ or $c$) and each set from $B$ that contains $x$, there is a subset that is in $B1'$ and also contains $x$ (namely $\{x\}$). It is true that $B1'$ also has some elements, like $\{a,c\}$ that are not needed for this verification, but that makes no difference. Once $B1'$ has enough elements to satisfy the criterion in the lemma, having some extra elements does no harm.
Jun
21
comment If $f\colon X\to Y$ is injective there is a $g\colon Y\to X$ such that $g\circ f=Id_X$
Both this answer and the question tacitly assume that $X\neq\varnothing$. If $X$ is empty and $Y$ isn't, then the result is false.
Jun
21
answered May two or more bases of different sizes generate a same topology?
Jun
21
comment Problems with this reasoning (Gambling)
I think this is a legitimate proof of the fact that, if you already have infinite wealth then you can win a dollar. Of course, I can imagine some more profitable things to do with your infinite wealth, and I can't imagine why you'd want more.