Suppose we add the axiom to ZF that for any set $S$, and any partition $P$ of $S$, the cardinality of $P$ is less than or equal to the cardinality of $S$. It is known that the axiom of choice implies that axiom. My question is, is the converse true? Or are there models of ZF where that axiom holds but the axiom of choice doesn't?
Here's a corollary of AC stating that: if there exists an onto mapping $f:A\to B$, then there exists a 1-1 mapping $g:B\to A$. What I'm confused about is that whether or not  AC can be derived from this proposition, in return? I have tried but haven't found a proof. Any help would be appreciated.
Is this notification that keeps coming up on my phone a virus? This notification keeps coming up, and i'm not sure if it's real or a virus, or how to stop it coming up. It says 'Android system warning! Google has found serious bugs in your Battery Module!Click on [OK] to view details.'   Screenshot (click for larger variant)  Any help appreciated, thanks. Lauren. Suppose that $f:[0,\infty)\rightarrow\mathbb{R}$ is continuous. Is it true that   $$\int_{0}^\infty|f(t)|dt&lt;\infty\Rightarrow \lim_{t\rightarrow\infty}f(t)=0?$$  If so can you provide a proof, otherwise a counter example. Thank you. Use Cauchy product to find a power series represenitation of   $$1 \over {(1-x)^3}$$ which is valid in the interval $(-1,1)$.   Is it right to use the product of $1 \over {1-x}$ and $1 \over {(1-x)^2}$ if I know thier expantion (HOW).. Is there another way? What are these cartridges that many characters have in the Star Wars movies? Admiral Ozzel has them on both sides of his uniform (upper chest).  Most Imperial officers seem to have them.  Han, pictured also here, has three such cartridges on his jacket.  I'd like to make such a jacket and use anything that looks like them.  I figure they are most closely related to cartridges for guns.  (which is what I imagine they are in the films).  But does anyone know the real scoop on them?  What is their purpose in the Star Wars universe?  Admiral Ozzel:    And Han Solo: Prove that a set of connectives is inadequate NO2 CSV downloaded files shows zero values for consider date range using Google Earth Engine When I am downloading the CSV file, the results show zeros in the CSV file. How can I download exact NO2 values as shown in Chart after running the mentioned below code in Google Earth Engine? // Create a FeatureCollection from the list and print it. var BergamoGeometry = ee.FeatureCollection(BergamoGeometry); print(BergamoGeometry); Map.centerObject(BergamoGeometry); Map.addLayer(BergamoGeometry); //////////////////////////////////////  var dataset = ee.ImageCollection(&quot;COPERNICUS/S5P/NRTI/L3_NO2&quot;)                  .filterBounds(BergamoGeometry)                   .filter(ee.Filter.date('2019-03-30', '2019-04-29'));                   var NO2 = dataset.select('NO2_column_number_density'); var band_viz = {   min: 0,   max: 0.0002,   palette: ['black', 'blue', 'purple', 'cyan', 'green', 'yellow', 'red'] }; // Create and print the chart. print(ui.Chart.image.series(NO2, BergamoGeometry, ee.Reducer.mean(), 30)); $a _{1} = \frac{1}{2}, a _{2} = 1, a _{n}= \frac{1}{2} a_{n-1} + \sqrt{a _{n-2} }$  Show by the induction that it's $\le 4$ and also by the induction that $a_{n+1}-a_n \ge 0$  If it comes to first I don't know what to do and if it comes to second I tried many different ways but I couldn't prove it. I won a Samsung Galaxy S6 Edge, signed into it with my Google account, played around with it, wiped it, and sold it.  Now the person can't get past setup because it's saying he must login with an account previously synced to the phone.   Is there anything I can do from my side to remove this lock? VB6 compiling via console (vb.exe) compatibility issue on 64bit (works on 32bit?) If $ \operatorname{Tr}(M^k) = \operatorname{Tr}(N^k)\;\forall 1\leq k \leq n$, then how do we show the $M$ and $N$ have the same eigenvalues? XML serialization without null XMLArray I was having a time getting my bootable USB 3, 32GB W10 install USB to be recognized. It recognized one time but I had to restart for something and it never came up again. I eventually removed all other boot up options to make it easier and faster to try again (no F12 needed).  Finally I gave up and tried another drive (16GB this time FWIW) and then it worked.  HOWEVER, it's stuck on a reboot loop now. I can choose to install, then to select the right disk, then it says it's loading file and then next thing I know it's saying do you want to install or upgrade. Is it normal to get hundreds of break-in attempts per day? I just checked my server's /var/log/auth.log and found that I'm getting over 500 failed password/break-in attempt notifications per day! My site is small, and its URL is obscure. Is this normal? Should I be taking any measures? How to get the filename without the extension from a path in Python? How to get the filename without the extension from a path in Python? For instance, if I had &quot;/path/to/some/file.txt&quot;, I would want &quot;file&quot;. How to execute .sh file on Windows? When I am trying to execute a file(name.sh) in the command line by the command  ./name.sh  , I am getting the error that:     "." is not recognized as an internal   or external command, operable or batch   file   please help me execute the .sh file Let $Q_8$ be the quaternion group. Show that $G/Z$ can never be isomorphic to $Q_8$, where $Z$ is the center of $G$. Hint: if $G/Z\cong Q_8$, show that $G$ has two abelian subgroups of index $2$.  I'm trying to prove the hint by looking at the canonical homomorphism $\pi:G\rightarrow G/Z$. Well, $Q_8$ has some abelian subgroups of index $2$, such as $H=\{1,-1,i,-i\}$, but that doesn't seem to map to abelian subgroups of index $2$ of $G$. (We know that $\pi(ab)=\pi(a)\pi(b)=\pi(b)\pi(a)=\pi(ba)$ for $a,b$ in $\pi^{-1}(H)$, but that only yields $aba^{-1}b^{-1}\in Z$, not $aba^{-1}b^{-1}=1$.) What is the number of automorphisms (including identity) for permutation group $S_3$ on 3 letters? Word for a person being used I'm looking for a word to describe someone who is being used. I want a noun to describe this person, not a verb or or adjective.  Maybe like a pushover. Since  and  you can get into a situation where your max HP is zero due to leveling up… what happens next?  What happens when your max HP is 0 because you leveled up with a negative Constitution modifier? Are you dead, unconscious, or what, and where do the rules say this? (The  assumes death, but that isn't established as fact anywhere yet.) Need help identifying a comic book about a wizard I need help with identifying the comicbook, what follows is my best effort to gather whatever I remember from reading it years ago (circa year 2000, but I'm positive it was older than that):   The main character was a guy with white hair, who wanted to become powerful.  Through some training and a deal of some kind he was able to become a mage of sorts. The training was set on some other world, or possibly even parts were happening in different dimension. His powers might have had something with travelling between dimensions, but I'm not  100% sure. Also, I seem to recall there was a hard limit of how many times he could use them. Then followed a part where he was famous, but someone framed him into a scandalous plot (it was a sexual scandal, I think) using his clone or a doppelganger. At the end of the issue     the main character is seen trapped in some strange state, unable to move or escape using his powers (this might have to do something with the deal he made earlier). I remember the scene at the end as this: he is suspended in mid-air, his friends look at him in disbelief while some malevolent figure explains to them that he is now trapped forever.  His name was the title of the comic. The second and third parts were set on futuristic-looking planet.   Rings any bells? I'd like to read it again only I can't google it out.