4,136 reputation
21647
bio website bruno.stonek.com
location Montevideo, Uruguay
age 25
visits member for 3 years, 10 months
seen 3 hours ago

Math student in Université Paris 13.


Jul
17
asked Interpretation of a formula and truth
Jul
17
comment Evaluating matrix-continued fractions?
Why not ask on MathOverflow?
Jul
16
comment Why some people don't like proofs by contradiction
I don't think I can give a good answer; while you wait for one, you should see read on intuitionistic logic (en.wikipedia.org/wiki/Intuitionistic_logic)
Jul
16
comment Solving $\lim\limits_{x \to 0^+} \frac{\ln[\cos(x)]}{x}$
@Javier: Exactly! You try to see your limit as the derivative of a function on a given point. Once you got it that way, you calculate the derivative of the recognized function using standard methods, evaluate it on the point, and voilà.
Jul
16
comment Solving $\lim\limits_{x \to 0^+} \frac{\ln[\cos(x)]}{x}$
@Javier: what's the definition of the derivative?
Jul
15
awarded  Nice Answer
Jul
15
awarded  Enlightened
Jul
15
comment What does “a map is isomorphic to another map” mean?
@Qiaochu: why can't/don't they introduce the xymatrix package to the SE network?
Jul
14
comment Can the indexing set of a cartesian product have the cardinality of the continuum?
@Theo: thank you!
Jul
14
comment Can the indexing set of a cartesian product have the cardinality of the continuum?
I suggest you take a look at en.wikipedia.org/wiki/Zermelo%E2%80%93Fraenkel_set_theory
Jul
14
comment Can the indexing set of a cartesian product have the cardinality of the continuum?
@Manos: great question. How do you know it? Axioms! For example, $\mathbb{N}$ is a set (axiom of infinity). Also, the power set of any set is a set (axiom of power set). A subset of a set is a set (axiom of comprehension). The range of a function is a set (axiom of replacement). (I'm being deliberately non-formal). You can prove from the usual axioms of ZF that $\emptyset$ is a set. So, we know there are some things which are sets by axioms, and then we know how to build more sets from them, by axioms which allow us some constructions.
Jul
14
revised Can the indexing set of a cartesian product have the cardinality of the continuum?
added 1 characters in body
Jul
14
comment Can the indexing set of a cartesian product have the cardinality of the continuum?
@Asaf: indeed, I added it for clarification, thank you.
Jul
14
revised Can the indexing set of a cartesian product have the cardinality of the continuum?
added 1063 characters in body
Jul
14
comment Can the indexing set of a cartesian product have the cardinality of the continuum?
@Manos: I'll add on the non-emptyness of the product to the answer, as per Asaf's suggestion.
Jul
14
comment Can the indexing set of a cartesian product have the cardinality of the continuum?
@Manos: No, I really meant exists. Not everything you define naïvely is a set ("the set of all sets" is not a set! Russell's paradox), you actually have to prove from the very axioms of set theory that you have defined something which really is a set... For a classical example, if $A$ and $B$ are sets, then as I said above, $A\times B:= \{(a,b):a\in A,b\in B\}$ is a set, but you have to prove it. If you describe a set by $\{x:\varphi(x)\}$ where $\varphi$ is a formula, you have to prove it exists: it may not exist (again, Russell's paradox: $\{x: x\notin x\}$ is not a set).
Jul
14
comment Can the indexing set of a cartesian product have the cardinality of the continuum?
@Manos: you're welcome. I added some formality to the answer.
Jul
14
revised Can the indexing set of a cartesian product have the cardinality of the continuum?
added 645 characters in body
Jul
14
answered Can the indexing set of a cartesian product have the cardinality of the continuum?
Jul
14
comment How to add compound fractions?
I'd like to point out, that, however, $x= \frac{1}{\frac{1}{x}}$ only when $x\not=0$.