Rudy the Reindeer
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 Mar 18 comment Continuous function on a compact metric space is uniformly continuous @Theo Oh, I see now what you mean! All this time I thought people where saying the $\delta$ were off by a factor of $2$. Duh. Of course one has to replace $\varepsilon$ by ${\varepsilon \over 2}$. The proof stays the same though. Mar 18 comment Continuous function on a compact metric space is uniformly continuous @Theo No: In the second paragraph the $\delta_i$ are actually equal to ${\delta_i \over 2}$ from the first paragraph. I believe this to be the factor of two that everybody is criticising in the comments. Right, that's the factor you are talking about? Mar 7 comment A polynomial that is zero on an open set This is not clear to me: Consider for example the polynomial $p(x,y) = xy$. Then this has degree one but $|\mathbb R|$ many roots. What am I missing? Feb 29 comment Munkres' Topology Problem @GiuseppeNegro Thanks for the link! Feb 28 comment Munkres' Topology Problem @DanielFischer I feel like a year ago I could do proofs like this with my eyes closed. Then I spent a year not doing any topology and now I can't even do the most basic thing. How depressing. Maybe my brain is broken. Feb 28 comment Munkres' Topology Problem @DanielFischer I deleted my shitey "answer". Feb 28 comment Munkres' Topology Problem @DanielFischer Hmmm. I am missing something. Why can the intersection not be closed while $A_\alpha$ aren't? Feb 19 comment Closed set via addition of two sets Slightly more interesting would be to show that $A+B$ is closed when $B = \{(x,x) \mid x \in R\}$. Feb 9 comment Find the matrix representing T and Find the Image of T (as a span of vectors) The matrix you found is correct. The image of $T$ is the span of the columns of $A$. Feb 5 comment Check the proof of $||x||^2$ is not a norm @ElChapo Her question is "Can someone check my proof for correctness" Feb 5 comment Check the proof of $||x||^2$ is not a norm Yes, that's all correct. You didn't have to do the third case: after 2) you already know that it is not a norm. Jan 30 comment A question about a proof of Noetherian modules and exact sequences I don't think a proof-verification can be a duplicate of a newer question. Unless someone posts the exact same proof with the exact same mistakes. Jan 22 comment Is the space $C[0,1]$ complete? @karhas Yes, that's right: it's definitely needed in the proof that $f$ is the uniform limit of the $f_n$. Jan 10 comment $a\mid b$ and $b\mid a$ but $a$ and $b$ are not associates I couldn't find a definition of associates (other what's in @frogeyedpeas' link and there it's the same as $a\mid b$ and $b \mid a$). Can you include a definition in your question? Jan 9 comment Continuous function on a compact metric space is uniformly continuous @user46944 No, it's not a typo. Jan 1 comment $f \in L^1 ((0,1))$, decreasing on $(0,1)$ implies $x f(x)\rightarrow 0$ as $x \rightarrow 0$ Oh, I wasn't trying to make a point -- I was merely asking a question (to which I don't know the answer). Jan 1 comment $f \in L^1 ((0,1))$, decreasing on $(0,1)$ implies $x f(x)\rightarrow 0$ as $x \rightarrow 0$ +1 by the way, for posting your solution, regardless of whether it's correct or not. Jan 1 comment $f \in L^1 ((0,1))$, decreasing on $(0,1)$ implies $x f(x)\rightarrow 0$ as $x \rightarrow 0$ Isn't $L^1$ the space of Lebesgue integrable functions? If it is, could you justify why you can use the Riemann integral in your first equation (the one with the $\infty$ on the left hand side)? Dec 29 comment Is this contradiction faulty? @ReinhildVanRosenú It seems to be a very confused person who wrote that website. For example, if $S$ is all the values where $f$ is $\le 0$ the person makes a case distinction for values where $f$ is $>0$. Perhaps you can find better proofs in a book. Or on Wikipedia. Dec 22 comment Modulo of a negative number But then your answer states that method 1 is correct while method 2 is not. Or is it not?