Set theory question about sequences

[bitter quail]

does anyone have any tips on how to solve 34?

[broken gorge]

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[little epoch]

cantor-bernstein

I suggest you prove that for infinite cardinal k it holds that k^2 = k

then it follows

$$ 2^\lambda \leqslant \kappa ^\lambda \leqslant (2^\lambda)^\lambda = 2^{\lambda ^2} = 2^\lambda $$

[kindred ginkgo]

aL

[little epoch]

@bitter quail

[bitter quail]

sorry i got the wrong one i meant 33 for 34 i solved it in two ways

thank you for helping me with 34

but do you have any ideas for 33

[little epoch]

what is Sq(A)?

set of sequences of elements of A?

[bitter quail]

?

[little epoch]

oh no, nevermind

$$ \mathrm{Sq}(A) \equiv \bigcup _{n\in\mathbb N} A^n $$

[kindred ginkgo]

aL

[little epoch]

all the components are equivalent to A because they are finite direct products

hence countable union of them is also equivalent to A

@bitter quail

[bitter quail]

yeah but how do you prove that

[little epoch]

just reproduce the proof that all finite sequences of natural numbers is countable

start with

$$ A^n \approx A $$

[kindred ginkgo]

aL

[bitter quail]

but we arent garanteed that A is countable

A could be uncountable

[little epoch]

that's not a problem

the proof is the same

[bitter quail]

oh?

[little epoch]

just write it out

[bitter quail]

yeah give me a min ill check that out

[little epoch]

$$ |A^n| = |A|^n = |A| $$

[kindred ginkgo]

aL

[little epoch]

to get you started

[little epoch]

owing to this fact

[bitter quail]

just to confirm you mean this proof right?

wait wrong one

[little epoch]

in your case the claim is countable union of cardinality X is still cardinality X

but the technique is the same

if it helps you can assume without loss of generality the union is disjoint

[bitter quail]

would a valid proof go along the lines of

[bitter quail]

we know

[bitter quail]

hence we have

[bitter quail]

so then

[bitter quail]

hence

is that right?

or am i missng smt?

[little epoch]

correct so far

now to take countable union of cardinalities |A|

i gtg to work, you're almost there

[bitter quail]

thank you

[little epoch]

yes, this is the correct one, it's probably given with diagonal argument in N

think of how to generalise this to cardinality |A|

[bitter quail]

@little epoch

i made it a bit better

[bitter quail]

.closed

+closed

[little epoch]

you make claims but no proofs