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21:00
and thats true
Thanks!!! @Jakobian @leslietownes
I just ask one last thing
columbo!
one of my favourite shows
why doesn't a symmetrical relation imply that it is also asymmetrical?
A relation $R$ is symmetric if $xRy$ implies $yRx$. Its asymmetric when $xRy$ implies that $\lnot yRx$
If there exists $x, y$ such that $xRy$ and $R$ is symmetric, then $R$ cannot be asymmetric, because that would imply that $yRx$ is not true. But its true, since its symmetric
That said, $R = \emptyset$ is an example of relation thats both symmetric and asymmetric
21:19
asymmetric if: $xRy$ and $yRx$ $\Rightarrow x = y;$
I see, I was using the wikipedia definition
In this case, if $xRy$ and $R$ is both symmetric and asymmetric (anti-symmetric on wikipedia), then $yRx$ and so $x = y$
So if $R$ is both symmetric and anti-symmetric, then $R\subseteq \Delta_A$
and conversely, if $R\subseteq \Delta_A$ then $R$ is both symmetric and asymmetric
So examples of such relations exist, and there is more of them than using previous definition
one example is equality like above i.e. $xRy$ iff $x = y$
Could you give me two examples (according to the definition I sent), one that imply both and of one which does not imply the other
pls
what do you mean?
by imply both/does not imply the other
21:37
symmetric if: $xRy$ $\Rightarrow yRx;$ asymmetric if: $xRy$ and $yRx$ $\Rightarrow x = y;$
Anyway, sorry if I reply after a while, but I'm on the phone, not on the PC :(
I know that those are your definitions, its just not clear what properties do you expect the examples to have
Jakobian was asking clarification about your question about the two examples
I think you should come up with those examples yourself
as Jakobian has already explained, most interesting relations aren't both symmetric and antisymmetric
so if you can produce interesting examples of symmetric and anti-symmetric relations, you will also have plenty of counter-examples
I didn't quite understand the difference between the two definitions
(the ones I sent 10 minutes ago)
symmetric if: $xRy$ $\Rightarrow yRx;$ asymmetric if: $xRy$ and $yRx$ $\Rightarrow x = y;$
lets say you have a relation on a set $A = \{1, 2, 3, ..., n\}$
Then you can describe a relation $R$ on $A$ in terms of a matrix having entries of $0$ and $1$
$B_{ij} = 1$ if $iRj$ and $B_{ij} = 0$ otherwise
21:49
they say different things, so they are different definitions
what does it mean for a relation $R$ on $A$ to be symmetric/asymmetric, in this example? In terms of the matrix $B = (B_{ij})_{i, j = 1}^n$
again, I suggest thinking about examples
consider something intuitive like a set $S$ consisting of the members of a family and let $R$ be the relation $xRy$ iff $x$ is an ancestor of $y$. is this symmetric? is it antisymmetric?
what does being an ancestor mean?
I think these are a bit advanced for Pizza
@Jakobian you know
21:52
I don't actually know what it means
oh you mean like a real life family
not a family of sets
yeah lol
@pizza try to find an example of a relation that is antysimmetric but it's not symmetric
@SineoftheTime I think people like to think in terms of matrices
@Thorgott The first time I read that, I read the last word as "antisemitic".
I don't know he's familiar with matrices, since he's at the beginning of a linear algebra course
21:57
@XanderHenderson oof, I prefer the "family of sets" misunderstanding
@XanderHenderson is it an antisemitic family? This will be on the test
22:08
0
A: Second countability of compact open topology

Dannyu NDosTo be concrete, here's my answer: If $X$ and $Y$ are second-countable and if $X$ is locally compact Hausdorff, $C(X,Y)$ is second-countable. Proof: Being second-countable locally compact Hausdorff, $X$ admits a countable basis whose every basic open set has compact closure; let $\mathcal{U}$ be o...

Any feedbacks?
@DannyuNDos X admits a countable basis whose every basic open set has compact closure
Thanks everyone, I'll see better tomorrow, because I don't understand much on the phone
this is true, but can you justify it?
I cant see the $$
@Jakobian Edited.
22:14
besides that I'm not sure if this question really needed another answer, the argument seems incomplete
you're just showing a basic open in one contains a basic open in the other, but you need to show it is a union of such
@Thorgott Munkres' Topology, p. 79.
Lemma 13.3.
I prefer communicating in words than with posting numbers from some book
I know what you're trying to do and I'm saying you didn't quite do it correctly
Yeah, you need to take $f\in \bigcap_{i=1}^n B_i$
and then show there is open $U$ in the other topology such that $f\in U\subseteq ...$
yeah, you need to show that every such element is contained in a basic open of $\mathcal{F}^{\prime}$
so what is missing is this taking of some arbitrary element
alternatively you can be lazy and cite Dugundji
22:21
well, the question already has an answer with a citation
no one actually gave any arguments so I think its fine
it was more of a discussion
22:33
Edited.
23:00
@DannyuNDos you wrote that $V_{ij}$ is contained by $V_i$. What does that mean?
23:23
Quite literally, $V_{i,j} \subset V_i$.
23:43
@DannyuNDos yeah this is wrong
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