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 May 10 comment Why does $\exists x\,\ x = x$? @RickyDemer So, basically, the difference between your formalization and the one in Wikipedia is that there's an additional existential qualifier $\exists\emptyset$ in the Axiom of Infinity? May 7 comment Prove that an open interval and a closed interval are not homeomorphic Right, it should be the induced topology. What I'm saying is that it isn't clear that on the induced topology that the whole unit open interval is not a compact set. On the induced topology it is closed after all. May 7 comment Prove that an open interval and a closed interval are not homeomorphic Why is $(0,1)$ not compact? Its compactness needs to be evaluated relative to its induced topology, not some unrelated superset like $\mathbb{R}$. May 6 comment Prove that an open interval and a closed interval are not homeomorphic Didn't you make assumptions about how the image looks like? May 6 comment How come Stone-Weierstrass theorem does not imply that in a given interval every continuous function has a power series expansion? I'd also note that Stone's extension to the Weierstrass Polynomial Approximation theorem generalizes its conclusions beyond the scope of polynomials anyway. Any point-separating real continuous function algebra on any compact $T_2$ space suffices for the approximation (complex algebras must also be self-adjoint). May 6 answered Prove that an open interval and a closed interval are not homeomorphic 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. The continuity of the function $f$ is a big assumption which was not stated in the question. 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. If $M(x)$ is not continuous from the left, then $\exists \epsilon >0\forall\delta...$, which is different from what you said. Further, $M$ is monotone increasing, not decreasing. 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. How would you show that the Heaviside theta is continuous from the left at 1? It is not. 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.