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Consider the pairs $(n,k)\in\mathbb{N}^2$ with $0\leq k<2^n$, they form a sequence in lexicographic order. Consider now the functions $f_{n,k}$ which are defined as $0$ in $[0,\frac{k}{2^n}]$, $1$ in …

Let $f=(\phi,\psi):\Omega_1\to\Omega_2$. For every point $p\in\Omega_1$ consider the curve $\gamma:t\mapsto p+\varepsilon e^{it}$, for $\varepsilon$ so small that the curve is contained in $\Omega_1$. …

If $V$ is closed in $\mathcal{O}$, then $\tau^V$ is continuous at $x_0$. If $V$ is not closed, then it is not hard to find counterexamples (e.g. imagine $\mathcal{O}=\mathbb{R}^3$, $V$ is an open disk …

The answer to question $1$ is yes: we can suppose $t=0,\gamma(0)=(0,0)$ and $\gamma'(0)=(1,0)$ for the purposes of this question. Then as $\frac{||\gamma(x)||}{|x|}\to 1$ when $x\to0$, there is some $ …

Let $f:\mathbb{R}\to\mathbb{R}$ be a differentiable function such that the set $A:=\{x\in\mathbb{R};f'(x)\not\in\{1,-1\}\}$ has measure $0$. Does this imply that $f'$ is constant? Context: I was think …

Such a function doesn't exist for some choices of $a$. For notational purposes I will change the unit interval by $[-2,2]$. Consider a $C^\infty$ function $a:[-2,2]\to\mathbb{R}$ such that $a(0)=3$, $ …

Here is a way to do it without the axiom of choice, but it isn't a nice formula either. Consider a Cantor set $C\subseteq[0,1]$ with Hausdorff dimension $0$. Now consider a countable disjoint union $\ …