Activity for watchmakerâ€
| Type | On... | Excerpt | Status | Date |
|---|---|---|---|---|
| Comment | Post #294950 |
Morjesta Tommi,
it looked to me like you were saying the set would contain itself *as *one* element*, not just as a *subset*. Maybe I've got your wording wrong. (more) |
— | 10 months ago |
| Edit | Post #295007 |
Post edited: Tell you what I missed |
— | 10 months ago |
| Edit | Post #295007 |
Post edited: missing case |
— | 10 months ago |
| Edit | Post #295007 |
Post edited: Typos, more or less |
— | 10 months ago |
| Edit | Post #295007 |
Post edited: Explaininng about "collapse" |
— | 10 months ago |
| Comment | Post #294950 |
Terve Tommi,
"A set is also its own subset.", yes, but not as an *element* of itself.
For "sane" sets, $X\not\in X$, otherwise we are not talking about *sets* any more but about classes.
https://en.wikipedia.org/wiki/Georg_Cantor went to the lunar bin for thinking about that. (more) |
— | 10 months ago |
| Edit | Post #295007 | Initial revision | — | 10 months ago |
| Answer | — |
A: Are there other topologies on $\mathbb R$ that make it a topological field? This is only half of an answer but here we go: First step is to get rid of those binary functions $+ : X\times X \to X$ and $ : X\times X \to X$, and with it, the need to consider product topologies. And to separate algebra from topology. Name the topology as $\mathcal T$. Addition and ... (more) |
— | 10 months ago |
| Edit | Post #295005 | Initial revision | — | 10 months ago |
| Answer | — |
A: How to justify: for every integer $r$ in $[1, k-1]$, there is an integer $j$ in $[0, k-1]$, such that $r + j = k$ Looks like some kind of "brain training". Looks very basic, but basic things can be very tricky if posed in a way you are not used to think. The way to tackle those things is to transform what is given into a representation you are used to work with. Lets go: $r+j=k$ is equivalent to $j=... (more) |
— | 10 months ago |
| Comment | Post #293500 |
I can see two ways to approach it.
One is to look at $\mathbb R$ as a vector space over $\mathbb Q$ looking at it as a product of topological spaces and giving each one of the obvious topologies. Doesn't realy look like the answer you are looking for.
Second way is to describe all the topol... (more) |
— | 10 months ago |
| Comment | Post #294950 |
"The power set of the empty set has two subsets; the empty set is there, as is the power set itself."
I think that part of your answer is wrong. The power set is not an element of the power set. We would be in big trouble if it were so ... think about "the set of all sets that do not contain i... (more) |
— | 10 months ago |