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 Apr 25 awarded Yearling Mar 17 comment Continuous function on a compact metric space is uniformly continuous @RudytheReindeer Ryker is right, you are missing a factor of 2. You say in a comment that $\delta_{x_i}$ is such that $d(x,y)<\delta_{x_i}$ implies that $d(f(x),f(y))<\epsilon$. But that's not true. From continuity, $\delta_{x_i}$ is such that if both $x$ and $y$ are in the $B(\delta_{x_i},x_i)$, then $f(x)$ and $f(y)$ are both in the $B(\epsilon,f(x_i))$, which gives the distance between $f(x)$ and $f(y)$ at most $2\epsilon$, not $\epsilon$ . Aug 20 awarded Popular Question Apr 25 awarded Yearling Mar 16 comment Proof that $S$ is a partial order when it is the reflexive closure of a strict partial order You are welcome. Indeed, it is not mentioned in the statement, but in the first line of the proof :). Mar 16 comment Proof that $S$ is a partial order when it is the reflexive closure of a strict partial order Yes, it is, and that's exactly the content of Theorem 4.5.2. Mar 16 answered Proof that $S$ is a partial order when it is the reflexive closure of a strict partial order Mar 16 comment Branch cut choice for $(z+\frac{1}{z})^{\frac{1}{3}}$ Hm.... Finding branch like that means finding a branch $\log_\tau$ of $\log$ such that for any $|z|>2$ one has arg$(z+1/z)\neq\tau$. I don't think that's possible. Mar 16 comment The Banach–Mazur distance for finite-dimensional $\ell_p$ @BenWallis If I am not mistaken, the reason is that, regardless how large $k_n$ is, an $n$-dimensional subspace is going to be isometric with $l_2^n$, and the Banach-Mazur distance between $l_2^n$ and $l_p^n$ goes to $\infty$ when $p\neq 2$. Thus, you cannot find a bound that depends on $p$ alone. Mar 15 answered How do I change the order of integration of this integral? Mar 15 comment The Banach–Mazur distance for finite-dimensional $\ell_p$ Yes, the conjecture is false for $p\neq 2$. By the way, by Dvoretzky, any infinite dimensional $X$ contains, for every $n$, almost isometric copies of $l_2^n$. Feb 7 answered A Property about orthogonal projections Feb 7 comment Algebraic and orthogonal complements $Z$ is an algebraic complement of $Y\subset X$ if any $x\in X$ can be written uniquely as $x=y+z$, with $y\in Y$ and $z\in Z$. A subspace $Y$ has many algebraic complements, but they are all isomorphic. Being isomorphic means that they have the "same dimension", which is to say bases in any algebraic complement having the same cardinality. When we speak about orthogonal complement, it is assumed additional structure exists (an inner product space), and the orthogonal complement is defined in terms of the inner product. Feb 6 answered Algebraic and orthogonal complements Feb 1 asked Isomorphisms with invariant linearly independent dense subset. Feb 1 comment Every isomorphism on a separable Banach space has a completely invariant dense subset Thank you. Can you pick $D$ such that is linearly independent? Don't tell me is still trivial :) Feb 1 accepted Every isomorphism on a separable Banach space has a completely invariant dense subset Feb 1 comment Every isomorphism on a separable Banach space has a completely invariant dense subset Of course, I see now. Thanks! Feb 1 comment Every isomorphism on a separable Banach space has a completely invariant dense subset I can see that $T(D)\subset D$. But why equality? If you exclude $n=0$ (so no $E$ in $D$), why is still dense? Probably I am missing something easy :) Feb 1 comment Every isomorphism on a separable Banach space has a completely invariant dense subset @Fundamental You are right, it was trivial. Modified it to what actually I am trying to prove. Still trivial?