My teacher said (without explaining why) that because $R^{n}$ is finite dimensional then set of all linear functions (Dual space) is equal to the set of all continuous linear functions. Can someone explaining why this true?  Thank you.
From      Every linear function on a finite-dimensional space is continuous.   I was wondering what the domain and codomain of such linear function are?   Are they any two topological vector spaces (not necessarily the same), as along as the domain is finite-dimensional? Can the codomain be a different normed space (and may not be finite-dimensional)?  I asked this because I saw elsewhere the same statement except the domain is a finite-dimensional normed space, and am also not sure if the codomain can be a different normed space (and may not be finite-dimensional).  Thanks and regards!
I have a pretty simple table that needs to span multiple pages. This seems like a job for either the / package or the  package.   The voting on the answers to this  suggest that longtable is the preferred package, yet xtab is much newer and has been revised since that question was asked.  What are the differences between xtab and longtable and when is one preferable to the other? I don't understand why the interior of $\mathbb{Q}$ in $\mathbb{R}$ is empty, since, for every ball with the center being a rational number, given an $\epsilon&gt;0$, I can find an infinite sequence of rational numbers that approach this point. For example, take $\frac{1}{2}$. The sequence $\frac{1}{2}+\frac{1}{n}$ can be made as close as I want to the number $\frac{1}{2}$, therefore I can always have open balls with center $\frac{1}{2}$ such that there are rationals inside it. Hamiltonian Monte Carlo for dummies Could you provide a step-by-step for dummies explanation of how Hamiltonian Monte Carlo work?  PS: I've already read the answers here, , and here, , and here,  and they do not address it in a step-by-step way. Interpretation of log transformed predictor and/or response I'm wondering if it makes a difference in interpretation whether only the dependent, both the dependent and independent, or only the independent variables are log transformed.  Consider the case of  log(DV) = Intercept + B1*IV + Error    I can interpret the IV as the percent increase but how does this change when I have   log(DV) = Intercept + B1*log(IV) + Error   or when I have   DV = Intercept + B1*log(IV) + Error   ? Proving Binomial Identity without calculus How to establish the following identities without the help of calculus:  For positive integer $n, $ $$\sum_{1\le r\le n}\frac{(-1)^{r-1}\binom nr}r=\sum_{1\le r\le n}\frac1r $$  and  $$\sum_{0\le r\le n}\frac{(-1)^r\binom nr}{4r+1}=\frac{4^n\cdot  n!}{1\cdot5\cdot9\cdots(4n+1)}$$ What's the rationale for requiring that a field be a $\boldsymbol{non}$-$\boldsymbol{trivial}$ ring? The title pretty much says it all.  Of course, one answer (IMO unsatisfactory) to such questions goes something like "a definition is a definition, period."  In my experience, mathematical definitions are rarely completely arbitrary.  Therefore I figure there must be a good reason for insisting that a field be a non-trivial ring (among other things), but it's not obvious to me.  I realize that a trivial ring would make for a very boring field, but it is also a rather boring ring, and yet it is not disallowed as such.  EDIT: I found a hint of a rationale in the statement that "the zero ring ... does not behave like a finite field," but I could not find in the  for this assertion exactly how the zero ring fails to behave like a finite field. Question about Angle-Preserving Operators This an exercise out of Spivak's "Calculus on Manifolds".  Edit:  There was a typo in the exercise as is noted below in the answers.  The statement has been edited to reflect this.  Given $x,y\in\mathbb{R}^{n}$, the angle between $x$ and $y$ is defined by  $$\angle(x,y) = \arccos\left(\frac{\langle x,y \rangle}{|x|\cdot |y|}\right),$$ where $\langle x,y \rangle$ denotes the standard Euclidean inner product.  A linear operator $T:\mathbb{R}^{n}\to\mathbb{R}^{n}$ is said to be angle-preserving if $\angle(T(x),T(y)) = \angle(x,y)$ for every $x,y\in\mathbb{R}^{n}$.  The exercise as stated:  Let $\{x_{1},\dots, x_{n}\}$ be a basis for $\mathbb{R}^{n}$.  Then suppose that $\lambda_{1}, \dots, \lambda_{n}\in \mathbb{R}$ are such that $Tx_{j} = \lambda_{j}x_{j}$ for each $j = 1,\dots, n$.  Then $T$ is angle-preserving only if (not if and only if!)$|\lambda_{i}| = |\lambda_{j}|$ for every $1\leq i\leq j\leq n$.  I'm having problems with the $(\Rightarrow)$ direction.  My best attempt (which seems to lead nowhere) is to suppose that $|\lambda_{j}|\neq |\lambda_{k}|$.  Then by assumption, \begin{align*} \angle(Tx_{j},Tx_{k}) &amp; = \arccos\left(\frac{\langle Tx_{j},Tx_{k} \rangle}{|Tx_{j}|\cdot |Tx_{k}|}\right)\\ &amp; = \arccos\left(\frac{\langle \lambda_{j}{x_{j}},\lambda_{k}{x_{k}} \rangle}{|\lambda_{j}{x_{j}}|\cdot |\lambda_{k}{x_{k}}|}\right)\\ &amp; = \arccos\left(\frac{\lambda_{j}\lambda_{k}\langle {x_{j}},{x_{k}} \rangle}{|\lambda_{j}|\cdot|\lambda_{k}|\cdot|{x_{j}}|\cdot |{x_{k}}|}\right)\\ &amp; = \arccos\left(\text{sign}(\lambda_{j})\text{sign}(\lambda_{k})\frac{\langle {x_{j}},{x_{k}} \rangle }{|{x_{j}}|\cdot |{x_{k}}|}\right)\\ \end{align*} may also be calculated as \begin{align*} \angle(Tx_{j},Tx_{k}) &amp; =  \angle(x_{j},x_{k})\\ &amp; = \arccos\left(\frac{\langle x_{j},x_{k} \rangle}{|x_{j}|\cdot |x_{k}|}\right). \end{align*}  Then since $\arccos$ is injective, I believe I can make the jump that $\text{sign}(\lambda_{j})\text{sign}(\lambda_{k}) = 1$, which does not resemble the conclusion that I should arrive at.  Note:  I wasn't sure what tag to put this under, so anyone who knows better please feel free to adjust.  Thanks for any help you can give. Why can't I use Stack Exchange if I'm under 13 years old, or if I'm under 16 years old and from the EU? Why Doesn't a Bootable USB Boot This question is asked often in Ask Ubuntu, sometimes with few hints about the situation.   Please provide a list of possible reasons to help troubleshoot the problem. How long does it take for police 'presence' to clear when they are alerted? I engaged with some enemy NPCs, and accidentally shot a civilian in the head (though not sure it would've mattered if I hit them somewhere else or not) as they ran across my screen. As a result of this, I was alerted by the game that the NCPD (the police) were automatically en route to 'investigate', and that I could avoid them by leaving the area and &quot;lying low for a while&quot;. How long does it take for the police to stop searching for me, and is there any kind of visual information on the HUD that alerts me to when NCPD is or is no longer looking for me? Why apt can't find the source code of any package? Short harvard citation? I need to cite the following way:  The first 3 Letters of the author + the year.  Example:     Kemper 2009 --> [Kem09]   But I don't have any clue how to do that! Incorrect plot using pgfplots (trigonometric functions like cos, sin and tan) No non-trivial clopen sets in $\mathbb{R}$? How to give a direct proof? How to give a direct proof of the following result?     Let $A$ be a subset of $\mathbb{R}$ such that $A$ is both open and closed. Then $A$ is either empty or all of $\mathbb{R}$?   My work:   If $A$ is not empty, let $u \in A$. Then there is some open interval $(a,b)$ such that $$u \in (a,b) \subset A.$$  Now if $\mathbb{R} - A$ is not empty either, then let $v \in \mathbb{R} - A$. Then there is some open interval $(c,d)$ such that  $$v \in (c,d) \subset \mathbb{R} - A.$$  Suppose that $u &lt; v$.   Since $$\emptyset \subset (a,b) \cap (c,d) \subset A \cap (\mathbb{R} - A) = \emptyset,$$ we must have $$(a,b) \cap (c,d) = \emptyset.$$ So we can conlclude that $$b \leq c.$$ But $b \in A$ and $c \in \mathbb{R} - A$. So we must have $$b &lt; c.$$  What next? Can anybody here please help complete the proof from here on?  An edit based on a comment by Marc Paul:   Let us assume that the set $A$ is a non-trivial clopen set in $\mathbb{R}$. Let us define a function $f \colon \mathbb{R} \to \mathbb{R}$ as follows:  $$f(x) \colon= \begin{cases} 1 \ \mbox{ for } \ x \in A; \\ 0 \ \mbox{ for } \ x \in \mathbb{R} - A. \end{cases}$$   Let $V$ be an open set in the range space $\mathbb{R}$. We show that the inverse image set $f^{-1}(V)$ is open in the domain space $\mathbb{R}$.   The following cases arise:   If $0, 1 \in V$, then $f^{-1}(V) = \mathbb{R}$.   If $0 \in V$ but $1 \not\in V$, then $f^{-1}(V) = \mathbb{R} - A$.   If $1 \in V$ but $0 \not\in V$, then $f^{-1}(V) = A$.   And, if $0 \not\in V$ and $1 \not\in V$, then $f^{-1}(V) = \emptyset$.   Thus, $f^{-1}(V)$ is open in the domain space $\mathbb{R}$.   Hence the function $f$ is continuous.   What next? How does this lead to our desired conclusion?   [Yet another edit, again based on valuable comments from Marc Paul: ]  So if both $A$ and $\mathbb{R} - A$ were non-empty, then let's suppose $a \in A$ and $b \in \mathbb{R} - A$, and we can assume without any loss of generality that $a &lt; b$.   Then $f(a) = 1$ and $f(b) = 0$. So by the intermediate value theorem there is a real number $c \in (a,b)$ such that $f(c) = 1/2$, which is a contradiction because the image set of $f$ does not contain $1/2$. How can one differentiate between Dr. (PhD) and Dr. (MD or DO)? Why am I still getting a password prompt with ssh with public key authentication? GRUB and LILO both fail to install to NVMe hard disk when installing Debian Who wrote about an energy free space elevator? Do we use abstract algebra in real life? Do we use abstract algebra in real life?   Is there any application of abstract algebra in real life? I am not talking about mirrors, just a plain window made of glass like material. Would it be possible to allow light pass only in one direction but not the other?