T. Eskin
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 10h comment Topological space : Exercise in “Bounded subset and sequence limits” @kaithkolesidou In that case, what does $(a_n)(x_n)$ mean? Also, if $X$ is just a topological space, the whole idea of boundedness is not necessarily sensible. 2d comment Is a one-dimensional vector space orthogonal? What exactly do you mean by a subspace being orthogonal? Orthogonal to what? 2d comment Show that $\{1/n:n∈N\}∪\{0\}$ is compact The set you seem to be talking about is not $[0,\frac{1}{n})$ but instead $\{\frac{1}{n}:n\in\Bbb{N}\}\cup\{0\}$. Maybe you should also write what is your initial open cover that $G$ is a subcover of and why this would give you the freedom to make the claimed choice of open intervals in the second paragraph of your post. Apr 26 comment Is $(0, \infty)$ closed in $\mathbb{R}-0$? Not all closed sets are compact. And not all non-closed sets are open. Apr 25 comment Continuous function differentiable almost everywhere, show f' is measurable @corkol. composition of continuous functions Apr 22 comment open\closed and disjoint sets under R2 @havakok. The continuous functions are $(x,y)\mapsto y-\frac{1}{x}$ and $(x,y)\mapsto x$. Apr 22 comment open\closed and disjoint sets under R2 @havakok. If $X$ and $Y$ are two sets, $X\setminus Y$ is the set of elements that are in $X$ but not in $Y$. Apr 22 comment Find a solution to $z+e^{-z}=a$ where $a>1$. Numerical approximations allowed or only closed form solutions? Apr 21 comment Prove that dim(U + V ) ≤ dim U + dim V . @user6156388. Take $U=V$ for example. Then $U+V=U$ so $dim(U+V)=dim(U)<2dim(U)=dim(U)+dim(V)$. And $U$ can be any non-empty subspace of any non-empty vector space. As another non-trivial example, take $W=\Bbb{R}^{3}$ and $U$ a line that is inside a plane $V$. E.g. $V=\{(x,z,y):z=0\}$ and $U=\{(x,y,z):y=0,\;z=0\}$. Then $U+V=V$, so $dim(U+V)=dim(V)=2<3=dim(U)+dim(V)$. Apr 19 comment Prove that dim(U + V ) ≤ dim U + dim V . @user6156388. Correct. And there is no other case when it would occur. In that instance, obviously U and V don't have to share a basis vector given an arbitrary basis for both, but it must be the case that you can find a vector that is a basis vector for both U and V. Apr 18 comment Is a space metric on the positive real numbers not complete? Note that $\lim_{x\to\infty}\frac{1}{x}$ has to be interpreted in terms of the metric $d$ and not in terms of the usual metric on $\Bbb{R}$. Apr 13 comment Is this outer measure really regular? What you have showed is that $\overline{\mathbb{S}}=\{\emptyset,X\}$. Does this violate outer-regularity? Apr 13 comment Proving that a certain function is well-defined For $T$ to be well defined you need to show that $Tf$ is a continuously differentiable function for any continuous $f$. I.e. $Tf\in C^1[0,1]$ for any $f\in C[0,1]$. But this is implied by the fundamental theorem of calculus as the answers below indicate. Apr 5 comment Prove that the vector space is 3D @copper. Great! you're welcome. Apr 5 comment Prove that the vector space is 3D @copper. Look up the definition for dimension. The dimension of a vector space is defined as the number of linearly independent vectors that spans it. Apr 5 comment Prove that the vector space is 3D @copper if $ax+be^{x}+c\sin(x)=0$ in $C(\mathbb{R})$ means that $ax+be^{x}+c\sin(x)$ is equal to the function that is zero everywhere. So this equation holds for all $x\in\mathbb{R}$. Now you can go ahead and choose your favorite $x$ values and plug them in to conclude that we must have $a=b=c=0$. I gave you one choice of such $x$ but there are plenty of other options for this. Apr 5 comment Prove that the vector space is 3D @copper $span\{x,e^{x},\sin(x)\}=\{ax+be^{x}+c\sin(x):a,b,c\in\mathbb{R}\}=V$ by definition of span and $V$. Mar 29 comment Prove that a positive definite matrix has a positive determinant (without eigenvalues) What have you tried so far? Mar 23 comment homeomorphism from $\mathbb{R}^2$ to open unit disk. What is $f(0,0)$? Mar 13 comment Mean of a squared random variable Since the OP already expressed that he figured this out, I'll write it out here. Since $Var(X)=E(X^{2})-E(X)^{2}$ for any random variable $X$ regardless of the distribution, then there is a general rule, mainly $E(X^{2})=Var(X)+E(X)^{2}$. I.e. if $X$ has mean $\mu$ and variance $\sigma^{2}$, then $E(X^{2})=\mu^{2}+\sigma^{2}$.