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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 $E(X^{2})=\mu^{2}+\sigma^{2}$.