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 May6 comment Prove that an open interval and a closed interval are not homeomorphic Didn't you make assumptions about how the image looks like? May6 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). May6 answered Prove that an open interval and a closed interval are not homeomorphic May3 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. May3 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. May3 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. May3 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? May3 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? May3 answered Show that $B$ represents an inner product. Apr30 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. Mar19 awarded Informed Mar6 awarded Notable Question Jan29 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 $\mathbb{N}$. Jan28 comment Why does $\exists x\,\ x = x$? OK - that answers everything, thanks! Jan28 accepted Why does $\exists x\,\ x = x$? Jan28 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. Jan28 comment Why does $\exists x\,\ x = x$? @AsafKaragila I have found the answer to my specific question directly in the first link, thanks. Jan28 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. Jan28 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! Jan28 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.