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May
3
comment Show that the functions $m(x) = \inf_{a\leq \xi \leq x}{f(a)}$ and $M(x) = \sup_{a\leq \eta \leq x} {f(\eta)}$ are both continuous from left.
But do you agree that by your definitions for the Heaviside function $M(x)=0$ whenever $x<0$ and $M(1)=1$, so we do not have left-continuity, so the question is ill-formed?
May
3
comment Show that the functions $m(x) = \inf_{a\leq \xi \leq x}{f(a)}$ and $M(x) = \sup_{a\leq \eta \leq x} {f(\eta)}$ are both continuous from left.
Take $f$ to be the Heaviside Theta on $[-1,1]$, which is 0 for $x <0$ but $1$ for $x\ge 0$, both defined and bounded. It seems that $M(x)=f$, but is clearly not continuous from the left at 1. Perhaps you have flipped the directions?
May
3
answered Show that $B$ represents an inner product.
Apr
30
comment A problem on divisibility theorem
I do not find this to be a convincing reason why squares and cubes mod 7 all follow this pattern.
Mar
19
awarded  Informed
Mar
6
awarded  Notable Question
Jan
29
comment Why does $\exists x\,\ x = x$?
@Hurkyl Actually, it seems that this convention may be necessary to the avoid the possibility of an empty model. According to this formulation of the Axiom of Infinity, we must presuppose the existence of $\emptyset$ in order for it to be an element of $\mathbb{N}$.
Jan
28
comment Why does $\exists x\,\ x = x$?
OK - that answers everything, thanks!
Jan
28
accepted Why does $\exists x\,\ x = x$?
Jan
28
comment Why does $\exists x\,\ x = x$?
Thank you for the thorough answer, it does make sense. However, as I learned here, it seems that no convention is necessary. $\exists x\,\ x=x$ must be true, since its negation implies $\forall x\,\ x\neq x$, which is the opposite of an axiom of first-order logic.
Jan
28
comment Why does $\exists x\,\ x = x$?
@AsafKaragila I have found the answer to my specific question directly in the first link, thanks.
Jan
28
comment Why does $\exists x\,\ x = x$?
@CliveNewstead I apologize that my question may have not been clear - the empty set's existence can be proven from the axiom of specification and infinity, it's just that Wikipedia asserted that $\omega$ was not needed, and I was asking why.
Jan
28
comment Why does $\exists x\,\ x = x$?
@CliveNewstead I found what I was looking for in Andres Caicedo's answer here - all one has to do is negate the existence theorem and prove by contradiction!
Jan
28
comment Why does $\exists x\,\ x = x$?
@AsafKaragila thanks for the links. I'd argue that this isn't a duplicate however. I see how the existence of the empty set follows from the existence of a set, but my question isn't about the former.
Jan
28
comment Why does $\exists x\,\ x = x$?
@CliveNewstead I would agree that this would be true if I accept the empty set axiom, but I thought the point was that we didn't need one. Without it, $\exists x\,\ x = x$ wouldn't follow from the axiom $\forall x\,\ x = x$.
Jan
28
comment Why does $\exists x\,\ x = x$?
@CliveNewstead I don't see why "there exists a set $x$" is meaningless, but, even supposing it is, I still don't see why I can accept $\exists x\,\ x=x$.
Jan
28
asked Why does $\exists x\,\ x = x$?
Aug
30
awarded  Popular Question
Aug
22
awarded  Yearling
Aug
8
comment Is the vector cross product only defined for 3D?
I don't understand why we have to prove that a cross product exists for any given $d > r$ - for any set of vectors with size $r$, any matrix which has a column space defined by the basis of this set of vectors is guaranteed to have a rank of at most $r$, and so a null space of $d-r > 0$. Since this null space is always nontrivial, a vector satisfying the criteria for the cross product can always be found.