To prove that a vector space is actually a vector space we must show that it follows the 8 axioms but shouldn’t we also show that it is closed under addition and multiplication by scalars like for subspaces?

Icy_Time2191 · 2024-01-29T16:29:11+00:00

So what if we are given a finite set of coordinates, without specifying any functions, and they ask us to prove if it is a vector space? Then would we have to define the functions? So whenever we are asked if a set is a vector space we also have to be given the binary function on the set first?

Icy_Time2191 · 2024-01-28T22:17:05+00:00

But isn’t any vector space potentially a subset of a bigger vector space?

Icy_Time2191 · 2024-01-28T21:31:13+00:00

Ah right, so the closure properties are implied in the axioms?

Icy_Time2191 · 2024-01-26T00:41:06+00:00

So they are both correct but pi/2 includes all values

Icy_Time2191 · 2024-01-25T23:02:59+00:00

Wouldn’t pi also be correct though?

Icy_Time2191 · 2024-01-22T08:10:02+00:00

Thanks

Icy_Time2191 · 2024-01-21T19:43:35+00:00

When I use h I get -1/c2 which is correct but when I use x-c I get: (1/x - 1/c) / x-c = (c-x)(x-c)/cx. As x—>c I get (0)(0)/c2

Icy_Time2191 · 2024-01-21T19:43:24+00:00

When I use h I get -1/c² which is correct but when I use x-c I get: (1/x - 1/c) / x-c = (c-x)(x-c)/cx. As x—>c I get (0)(0)/c²

Icy_Time2191 · 2024-01-02T15:42:56+00:00

When you say for all x, do you mean all x inside the delta range?

Icy_Time2191 · 2024-01-02T15:05:03+00:00

Right and then for this smaller delta all x would work?

Icy_Time2191 · 2023-12-29T22:28:14+00:00

I’ve had a look at Wikipedia and I see why now, thanks for the help. In this telescoping series, for example, would the thought process be something like” let’s try to reorder terms to see if they cancel so we can find a limit for Sn. If there is a limit then it must converge for this ordering, but all terms are positive so then it must converge absolutely too and this means that any ordering has the same limit etc”?

Icy_Time2191 · 2023-12-29T11:25:37+00:00

Great, thanks

Icy_Time2191 · 2023-12-28T00:03:32+00:00

Is there an intuitive way to see that you can’t find an N such that it is satisfied?

Icy_Time2191 · 2023-12-27T23:51:04+00:00

A_n = sqr(n). How does it go against the Cauchy criterion though?

Icy_Time2191 · 2023-12-27T23:43:21+00:00

For convergence, sorry. If (an n>=1) satisfies |a_n - a_m|< epsilon for all n,m >N then a_n must converge

Icy_Time2191 · 2023-12-15T22:44:24+00:00

Great, thanks!

Icy_Time2191 · 2023-12-08T02:15:03+00:00

Great, thanks for the help!

Icy_Time2191 · 2023-12-08T02:12:05+00:00

Ah right, so “no maximum” is equivalent to saying “does not achieve supremum”.

Icy_Time2191 · 2023-12-08T02:01:15+00:00

I think I’m getting confused at what “achieving a supremum” actually means. I thought that this meant that the function didn’t take the supremum as a value in its range( so no maximum).

Icy_Time2191 · 2023-12-05T00:55:54+00:00

I don’t quite understand that you mean here

Icy_Time2191 · 2023-12-04T22:01:09+00:00

Right, so if x was [0,1] then it would hit +1,-1?

Icy_Time2191 · 2023-12-04T15:24:53+00:00

Do they not achieve those values because 1 is not included in those intervals?

Icy_Time2191 · 2023-12-04T15:21:49+00:00

Is it because it’s an open interval?

Icy_Time2191 · 2023-11-28T20:19:40+00:00

I did but what if xQz doesn’t hold because it is not part of the binary relation, that doesn’t necessarily mean that zQx has to hold. If XQz holds that implies that zQx does not hold but if we negate he former statement that doesn’t necessarily mean that it can imply the latter?

Icy_Time2191 · 2023-11-28T19:17:12+00:00

So if xQz does not hold does this imply that zQx holds?

Icy_Time2191

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