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11h
comment $(0,1) \simeq [0,1] $ but not homeomorphic
Any two contractible spaces are homotopy equivalent.
11h
comment $(0,1) \simeq [0,1] $ but not homeomorphic
What do you know about the word "contractible"?
11h
comment Power series converging at the convergence radius
This and this question are not entirely unrelated. The behaviour on the boundary of the disk of convergence can be complicated, even if the series converges at every point of the boundary.
12h
revised How to conclude this proof that real and imaginary parts of holomorphic functions are harmonic.
More descriptive title
13h
comment Is the odd part of even almost perfect numbers (other than the powers of two) not almost perfect?
Cool, thanks @Anonymous-agroup.
13h
comment Is the odd part of even almost perfect numbers (other than the powers of two) not almost perfect?
@Anonymous It's not bad to add appropriate tags to older questions, but at the rate at which you're going, you flood the front page with these edits. Can you slow down a bit, so that these edits only take up a reasonable portion of the front page, please?
15h
comment Is one-point compactification of a space metrizable
Yes and Yes. For 1., show that $Y$ has a countable basis.
17h
comment maximum frequencies of numbers in a matrix
Code Chef February long challenge question
2d
awarded  Nice Answer
Feb
6
awarded  Nice Answer
Feb
6
comment Dense convex set in $*$-weak topology
You need $\xi = x$. But you're looking at $(X',\sigma(X',X))$, and $(X',\sigma(X',X))' = X$.
Feb
6
comment An upper bound for the Chebyshev function?
Thus you absolutely cannot deduce $$\frac{\psi(x) - x}{\psi(x) - \theta(x)} < \frac{c_2 - 1}{c_2 - c_1}$$ just from the bounds $c_1 x < \psi(x) < c_2 x$ and $\frac{1}{2} c_1 x < \theta(x) < c_2 x$.
Feb
6
comment An upper bound for the Chebyshev function?
If you are looking at just one fixed $x$ ($\geqslant 4$), then the quotient is of course finite. But you cannot give a bound for it that is independent of $x$. Look at $a(x) = x+2$ and $b(x) = x + 1 + \frac{x}{1+x}\cdot \cos (\pi x)$. Then clearly $1 < \frac{a(x)}{x} < 3$ and $1 < \frac{b(x)}{x} < 3$ for $x > 1$, and $b(x) < a(x)$. But $$\frac{a(x) - x}{a(x) - b(x)} = \frac{2}{1 - \frac{x}{1+x}\cos (\pi x)}$$ becomes arbitrarily large, for $x = 2n,\, n \in \mathbb{N}\setminus \{0\}$, it is $\dfrac{2}{1-\frac{2n}{2n+1}} = 4n+2$.
Feb
6
comment Dense convex set in $*$-weak topology
You need $x\neq 0$, for $x = 0$ we have $\{f(x) : f\in W\} = \{0\}$. Use (one of) the Hahn-Banach separation theorem(s).
Feb
6
comment An upper bound for the Chebyshev function?
No. In $$\frac{\frac{\psi(x)}{x} - 1}{ \frac{\psi(x)}{x} - \frac{\theta(x)}{x}}$$ both, numerator and denominator tend to $0$ as $x\to \infty$. Thus we can't say that the fraction is bounded without knowing much more about the distribution of primes. And while it's straightforward to see that the denominator tends to $0$, the fact that the numerator tends to $0$ is equivalent to the PNT. So without the PNT, we couldn't rule out that the denominator is bounded below in absolute value by some $c > 0$, and then the (modulus of the) quotient would tend to $\infty$.
Feb
6
comment An upper bound for the Chebyshev function?
Concerning the addition to your comment, note that $$\lim_{x\to\infty} \frac{\psi(x)}{x} = 1 = \lim_{x\to\infty} \frac{\theta(x)}{x}.$$ This is equivalent to the prime number theorem.
Feb
6
comment An upper bound for the Chebyshev function?
If by that you mean $\sup_x \frac{\psi(x) - x}{\psi(x) - \theta(x)} < \infty$, then: No, it is not sufficient to establish that. Not nearly. The $\psi(x) < kx$ and $\theta(x) < kx$ doesn't even rule out $\psi(x) \geqslant \theta(x)$ infinitely often. We know that that doesn't happen for $x \geqslant 4$, but we cannot deduce that just from the bounds.
Feb
6
comment An upper bound for the Chebyshev function?
Okay, so you mean $\psi(x)/x$ and $\theta(x)/x$ are uniformly bounded above. But that doesn't give you much, it is possible that $\psi(x) - \theta(x)$ is quite small. It might be much smaller than $\psi(x) - x$. It's possible that that never happens, but we can't rule it out (yet), we don't know enough about the distribution of primes for that so far.
Feb
6
comment An upper bound for the Chebyshev function?
I don't understand that question. Neither $\psi(x)$ nor $\theta(x)$ is bounded above.
Feb
6
comment An upper bound for the Chebyshev function?
For a fixed $x \geqslant 4$, the quotient is of course finite since then $\psi(x) > \theta(x)$. But you can't deduce that the quotient is bounded above uniformly in $x$. If $\psi(x) - x > c\cdot x^{2/3}$ infinitely often for some $c > 0$, the quotient becomes arbitrarily large.