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Dec
20
revised Space of bounded continuous functions is complete
added 1 characters in body
Dec
20
asked Swapping a limit and a $\sup$
Dec
20
comment Space of bounded continuous functions is complete
@t.b. Sorry but I'm not sure I understand your first comment.
Dec
20
revised Space of bounded continuous functions is complete
added 74 characters in body
Dec
20
comment Space of bounded continuous functions is complete
@t.b.: Oh, right. In the question it says continuous. I'll do that. Thanks for your comments!
Dec
20
answered Space of bounded continuous functions is complete
Dec
20
comment Topology exercise from Manetti, Topologia
Hey stef. People around here like it when you accept some of their answers from time to time. To do so you click on the tick symbol next to the answer you'd like to accept.
Dec
19
revised Why is the maximal value attained at the boundary?
added 30 characters in body; edited title
Dec
19
answered understand a statement from a Book.
Dec
19
comment Where is the “relation” here?
@drozzy That was a typo. Here is the corrected version: No, the $\cdot$ does not denote function composition, it denotes the group operation. An example of a group where the group operation is denoted in a multiplicative way would be the reals without zero together with multiplication: $G = (\mathbb{R} \setminus \{ 0 \}, \cdot)$. Note that you have to exclude $0$ because it doesn't have an inverse and so $G$ would not be a group. For the monoid case it doesn't matter, $M = (\mathbb{R},\cdot)$ is a monoid.
Dec
19
comment Where is the “relation” here?
@sdcvvc True! Thanks for pointing it out.
Dec
19
comment Where is the “relation” here?
@drozzy Hope this helps, otherwise don't hesitate to ask for more explanation.
Dec
19
comment Where is the “relation” here?
(cont'd) where $e$ denotes the identity element of the group. In $\mathbb{Z}$ $e$ would be $0$ and in $\mathbb{R}$ $e$ would be $1$.
Dec
19
comment Where is the “relation” here?
An example of a group where you'd denote the operation using additive notation $+$ would be $G = (\mathbb{Z}, +)$. A group is like a monoid with one additional requirement: the requirement that every element must have an inverse i.e. for every $g \in G$ you can find a $-g$ (note that I'm using additive notation here) such that $g + (-g) = e = (-g) + g$.
Dec
19
comment Where is the “relation” here?
Hi @drozzy: No, the $\cdot$ is not a function decomposition, it denotes the group operation. An example of a group where the group operation is denoted in a multiplicative way would be the reals together with multiplication: $G = (\mathbb{R}, \cdot)$.
Dec
19
revised Where is the “relation” here?
edited tags
Dec
19
answered Where is the “relation” here?
Dec
19
answered Searching a binary search tree for a specific value
Dec
19
accepted Smooth functions with compact support are dense in $L^1$
Dec
18
comment Norm with special conditions
You can accept answers by clicking the tick symbol next to the answer you'd like to accept. People around here appreciate it if you accept one answer or the other from time to time.