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comment Example of Topological Vector Space
I think you need to narrow it down a little bit, because the whole space is obviously a convex neighbourhood of each of its points, so any non-locally convex space will satisfy your query.
Aug
17
awarded  Yearling
Aug
6
awarded  Populist
Aug
3
comment What is the meaning of $\exp(\,\cdot\,)$?
Wow. I just noticed that I have written "$\exp(a) + \exp(b)$" instead of $\exp(a) \exp(b)$. What a bizarre typo/matho.
Jul
19
comment Help hint on the following question regarding countable dense set and Lebesgue measure
@amirbd89 what's there to explain? Okay, using a minus sign for set difference might not be the best idea, but other than that it's pretty clear: It's the Lebesgue measure of the complement (in $\mathbb R^n$) of the union of translates of $E$ by elements of $D$.
May
4
comment Prove $\bigcap S$ exists for all $S \ne \emptyset$. Where is the assumption $S \ne \emptyset$ used in the proof?
For example if you instead wrote $\cap S = \{x \in \cup S \mathrel\colon A \in S \implies x \in A\}$, then there would be no problems and you'd have $\cap \emptyset = \emptyset$.
May
4
comment Prove $\bigcap S$ exists for all $S \ne \emptyset$. Where is the assumption $S \ne \emptyset$ used in the proof?
Writing $\{x : \Phi(x)\}$ for some formula $\Phi$ is dangerous (as per Russel's paradox). This is unrestricted set comprehension and is not allowed in ZFC.
May
1
revised What is the number of elements of $\mathbb Z[i] /I $, where $I:=\{a+ib \in \mathbb Z[i] : 2 \mid a-b\}$?
Use \mid for "divides"
May
1
revised What is the number of elements of $\mathbb Z[i] /I $, where $I:=\{a+ib \in \mathbb Z[i] : 2 \mid a-b\}$?
Cleanup and grammar fixes
Apr
20
comment Problem about $\sin(1/x)$ in topology. (open and closed functions)
I also do not understand your purported argument as to why $f$ is continuous. Isn't it much easier to argue that $f$ is the composition of continuous maps?
Apr
20
comment Problem about $\sin(1/x)$ in topology. (open and closed functions)
@AlexR: Exhibiting an open subset $A$ of $(0,\infty)$ such that $f(A) = [-1,1]$ does not show that $f$ isn't an open mapping since $[-1,1]$ is open in the codomain.
Apr
20
revised Problem about $\sin(1/x)$ in topology. (open and closed functions)
less-than-or-equals is not written as "<=" in (La)TeX
Apr
20
comment Is it true that if $g,h\in G$ have order $p$, where $p$ is prime, the only possible order of $\langle g\rangle \cap \langle h \rangle$ is $p$?
$$o(x) \mid \lvert \cdots \rvert$$ may not be as bad as the legendary $$\Xi \over \bar\Xi$$ but I do still think that it deserves a good old "your notation sucks!" May I suggest using $\#$ for "number of elements of" next time?
Apr
14
revised What's the Lebesgue measure of this set?
added 227 characters in body
Apr
14
answered What's the Lebesgue measure of this set?
Apr
10
answered Proving $\int_a^b f = - \int_b^a f$ from the reflection property of the integral?
Apr
7
comment Why are vector valued functions 'well-defined' when multivalued functions aren't?
It seems that you missed my point. I was saying that it isn't clear from $G$ whether I was talking about $x \mapsto x^2$ as a function from $\mathbb R$ to $\mathbb R$ or as a function from $\mathbb R$ to $[0,\infty)$. It's clearly only surjective in the latter case. Expanding the codomain of a function doesn't change its value, behaviour or graph (in analysis we most of the time don't really care about specifying the codomain too precisely, even), but it does change it as a function since it may go from being surjective to being... well... not.
Apr
6
comment Why are vector valued functions 'well-defined' when multivalued functions aren't?
Indeed, the Bourbakist definition of (partial) function is as an ordered triple $f = \langle D,G,C\rangle$ "(domain, graph, codomain)" such that $$G \subseteq D \times C \text{ and }(x,y),(x,y') \in G \implies y=y'.$$ A function is then a special case of partial function where $$x \in D \implies (\exists y \in C : (x,y) \in G).$$
Apr
6
comment Why are vector valued functions 'well-defined' when multivalued functions aren't?
If we're going to be pedantic, then the "a function is its graph" definition is just barely not good enough because it doesn't allow one to talk of surjective functions. For example we could look at $$G = \{(x,y) \in \mathbb R^2 \mid y = x^2\} = \{(x,y) \in \mathbb R \times [0,\infty) \mid y = x^2\}.$$ Is $G$ the graph of $\mathbb R \owns x \mapsto x^2 \in \mathbb R$, or of $\mathbb R \owns x \mapsto x^2 \in [0,\infty)$ or maybe it has some entirely different codomain?
Mar
30
revised Computing the Laplace transform of $\frac {f(x)}{x}$
added 312 characters in body