Inverse images of Borel sets by a continuous function are Borel Let $f\colon \Bbb R^p \to \Bbb R$ be a continuous function and $B\subseteq \Bbb R$ be a Borel set. Show that $f^{-1}(B)\subseteq \Bbb R^p$ is a Borel set.
Borel set preserved by continuous map Let $f:\mathbb{R}^m\rightarrow\mathbb{R}^n$ be a continuous map. Show that if $A$ is a Borel subset of $\mathbb{R}^n$, then $f^{-1}(A)$ is a Borel subset of $\mathbb{R}^m$.   I know that for $A$ open subset of $\mathbb{R}^n$, then $f^{-1}(A)$ is open. Likewise for closed. A Borel subset is a countable unions/intersections of these sets, done repeatedly. From those facts, combined with $f^{-1}(A\cup B)=f^{-1}(A)\cup f^{-1}(B)$ and $f^{-1}(A\cap B)=f^{-1}(A)\cap f^{-1}(B)$, the statement should be intuitively true.  But to do it rigorously, I want to write $A$ as open and closed sets. But since the countable union/intersection can be done repeatedly, I don't know how I can write it that way.
When is R squared negative? My understanding is that $R^2$ cannot be negative as it is the square of R. However I ran a simple linear regression in SPSS with a single independent variable and a dependent variable. My SPSS output give me a negative value for $R^2$. If I was to calculate this by hand from R then $R^2$ would be  positive. What has SPSS done to calculate this as negative?   R=-.395 R squared =-.156 B (un-standardized)=-1261.611   Code I've used:  DATASET ACTIVATE DataSet1.  REGRESSION /MISSING LISTWISE /STATISTICS COEFF OUTS R ANOVA             /CRITERIA=PIN(.05) POUT(.10) /NOORIGIN             /DEPENDENT valueP /METHOD=ENTER ageP   I get a negative value. Can anyone explain what this means? How to say the /p/ sound in the middle of words? I'm not sure how to say the /p/ sound in the middle of words like apple, puppy, happy etc. Those "p" sounds have a puff of air or not.  For myself I think those /p/ sounds have a puff of air but less than /p/ in the first of words like pen, pan etc. Is it correct or not?  However, I just listened a word "apple" from a video of Youtube but I can't hear or catch a puff of air from /p/ in apple.  It sounds like /p/ after /s/ like the words spy, spin etc. How does Windows 7 licensing work for running the OS as Virtual Machines? I plan on buying Windows 7 Ultimate Edition to run on an old Pentium 4. I also want to run Windows 7 virtual machines inside Windows 7. My question is two fold:   Do I need a license for each Virtual Machine? Does the answer change if I use a third party virtualization platform (e.g. , , , etc) as opposed to ? why __getitem__ cannot be classmethod? I'd like to create a Yahoo! account. The signup page won't let me proceed without entering a cell phone number, and it won't recognize my Google Voice number as a valid cell phone number. Is there a workaround to be able to create a Yahoo! account without having a cell phone that can receive SMS messages? Can we get x comment votes per y minutes, rather than one vote per 5 seconds? I quite often will read a thread of comments, and want to vote up a few of them. I vote one up, read the next few comments, vote up another and get a *giant red box*.  I've seen that "You can only submit a comment vote every 5 seconds" box far too often. Can I vote for 12 comments per minute, rather than 1 comment every 5 seconds? I'm having problems with an exercise from Apostol's Introduction to Analytic Number Theory.     Given $x$ and $y$, let $m=ax+by$, $n=cx+dy$, where $ad-bc= \pm 1$. Prove that $(m,n)=(x,y)$.   I've tried to give a proof, but I suspect it's wrong (or at least not very good). I would be very thankful for any hints/help/advice!  My proof:     We observe that since $ad-bc= \pm 1$, $ad=bc \pm 1$, and $(ad,bc)=1$.   Now, $(a,b) \mid m$ and $(c,d) \mid n$ but      $$(a,b) = ((d,1)a,(c,1)b)=(ad,a,bc,b)=(ad,bc,a,b)=((ad,bc),a,b)=(1,a,b)=1.$$      Similarly, we determine $(c,d)=1$. So, $1=(a,b) \mid m$ and $1=(c,d) &gt; \mid n$. But $(x,y)$ also divide $m$ and $n$. Since $(x,y) \geq (a,b)=(c,d)=1$, this implies that $(x,y)=m,n$. Hence   $(m,n)=((x,y),(x,y))=(x,y)$. For which $x$ exists sum $1 + 2 + 3 + \ldots + n$, where $n &gt; 3$, which has notation $xxx\ldots x$?   In other words, I am looking for a sum of natural numbers which gives a result which has all same digits, e.g. $5555555$ or $22222222222222222$, or whatever else.  How to find this series and how to proof it for all possible series?  I have found the following examples:   $1 + 2 + 3 + \ldots + 10 = 55$ $1 + 2 + 3 + \ldots + 11 = 66$ $1 + 2 + 3 + \ldots + 36 = 666$     And what about next series which ends with (different) repeating $x$? Hot to find and proof it? How can I make a view that seems like orthographic, but not exactly? For the start, I thought was orthographic, but when I used the orthographic camera, it didn't match up. Client say they want a so-called &quot;industry schematic&quot;, talking about a 135 degree look.  I found a trick to do this, which is using lattice. I align the lattice with the cube, and move the upper four points a little bit along the axis showing on the following image. But I want to achieve this by just using the camera. I am using ArcGIS Desktop 10.1.   I am trying to find the location of the highest value pixel in my Direct Solar Radiation Raster. I can see the extent of the data in my raster but don't know where the highest value pertains to.   Is there a way to simply select pixel with highest value or create a point at highest value? Proof that $\bigcap_{a\in A} \bigg(\, \bigcup_{b\in B} F_{a,b} \, \bigg) = \bigcup_{f\in ^AB} \bigg(\, \bigcap_{a \in A} F_{a,f(a)}\,\bigg) $ I found an interesting exercise of other amazing book (Jech's) that I'm not totally sure about how to do it.   Prove the following form of the distributive law:   $$\bigcap_{a\in A} \bigg(\, \bigcup_{b\in B} F_{a,b} \, \bigg) = \bigcup_{f\in ^AB} \bigg(\, \bigcap_{a \in A} F_{a,f(a)}\,\bigg) $$  Assuming that $F_{a,b_1} \cap F_{a,b_2} = \emptyset $ for each $a\in A$ and $b_1,b_2\in B$ and $b_1\not=b_2$.  First of all, I'm not completely sure about what $ F_{a,b}$ really means. I suppose that is  the range of a family with a domain $A \times B$.    And second, I cannot figure out how to do the converse (assuming that the first part it is correct, I used as normal in this kind of proof element chasing)  Proof:  ($\Rightarrow$) ...  ($\Leftarrow$) Suppose that $z\in \bigcup_{f\in ^AB} \big(\, \bigcap_{a \in A} F_{a,f(a)}\,\big)$; then there is $f\in \,^AB$ such that $z\in  \bigcap_{a \in A} F_{a,f(a)}\, $. Let $a \in A$ be arbitrary.   Claim 1 $\, \bigcap_{a \in A} F_{a,f(a)}\subseteq F_{a,f(a)} \subseteq \bigcup_{b\in B} F_{a,b}$.  Proof of the Claim 1:   For the first inclusion: Suppose $z\in\bigcap_{a \in A} F_{a,f(a)}$; then for each $a\in A$ we have that $z\in F_{a,f(a)}$. Then $\bigcap_{a \in A} F_{a,f(a)}\subseteq F_{a,f(a)}$.  For the second inclusion: Now suppose $z\in F_{a,f(a)}$. Clearly $f(a)\in B$ because $f\in ^AB$. Then there is some $b\in B$ such that  $z\in F_{a,b}$ and hence that $z\in \bigcup_{b\in B} F_{a,b}$. $\square$  Since $a$ was arbitrary it follows that $z\in \bigcup_{b\in B} F_{a,b}$ (by claim 1) for each $a\in A$ and hence  that $z\in \bigcap_{a\in A} \big(\, \bigcup_{b\in B} F_{a,b} \, \big)$ as desired.  I really, really would very much appreciate some help with that. Thanks in advance as usual. (clickable)  Mainad has a valid height and width, however it isn't shown like subad1/subad2. Which are in essence exactly the same! (just a different background image).  Firebug shows my div as greyed out for some weird reason. Replacing the contents of mainad with just some text doesn't solve the problem (problem isn't related to inner elements)..   What's going on?! :( How to fix slow wireless on machines with Intel wireless cards? Removing all fraction symbols like “¼” and “½” from a string $\sum a_{n}$ is convergent, $\sum a_{n}^2$ is divergent. Prove $\sum a_{n}$ is conditionally convergent. Suppose that the series $\sum_{n=1}^{\infty}a_{n}$ converges while $\sum_{n=1}^{\infty}a_{n}^2$ diverges. Prove that $\sum_{n=1}^{\infty}a_{n}$ converges conditionally. What are the pros and cons of the leading Java HTML parsers? Was Kylo Ren's Blaster Bolt Freezing Really Force Telekinesis? In this questions answer its stated that Kylo Ren freezes the blaster bolt in midair with telekinesis:       Because he is extremely skilled in Force telekinesis, Kylo Ren is able   to freeze not just objects but even blaster bolts in midair.   But as a commenter already pointed out, this seems nonsense (even in star wars force physics). The blaster bolt is, as far as I know, a cloud of plasma. If it is stopped by telekinesis, why should it continue its old path with its old speed after it is released? Wasn't it more a completely new force ablility like "time freeze"? While I admit in a case of time freeze the bolt should not have flickered all the time. Also the plasma should have expanded and vanished or simply cooled down releasing its energy to the surrounding air.  So when it was force telekinesis Kylo must have done the following:   Stop every atom in the plasma cloud Keep the cloud compressed and in form and shape. Isolate it to prevent the heat flow or keep it on temperature (is this still telekinesis?) Not just releasing it but accelerate it again exactly on its old path and speed.   This would be indeed a greate never seen mastery of telekinesis. And very ineffective. And if he wanted to be cool, wouldn't it be even cooler to change the path after the "release" so that it hits one of his enemies?  So what was going on there? Pure telekinesis or something else? Did the writers intend this? Or did they just not thinking it to the end? Is there a way to set transparency of windows in GNOME? How to draw a Sankey Diagram using TikZ Generating random integer from a range