ok, here it goes... Critique me, please. and keep in mind that I'm new at this.
"Prove that the set A of all sequences consisting only of zeros and ones has the power of the continuum."
To show that
has the power of the continuum, we will establish a one-to-one correspondence between
and
. First consider two subsets of
, which we will call
and
.
consists of all sequences having a finite number of zeros.
consists of all sequences having an infinite number of zeros.
The subset
is countable. Take a countable subset of
, say
, which we will call
. We can form a bijection between this set and
, call it
.
Now rewrite every sequence in
as a number written in base 2, keeping the number and order of the ones and zeros the same. Clearly there is a one-to-one correspondence between these numbers and
.
Rewrite
as
and
as
. Define our bijection as
whenever a sequence
and as rewriting the terms of
as a number in base 2 whenever
.
Since any subset of
is equivalent (denoted
) to the interval
(that is, it has the power of the continuum), we have that
. So, we have
and
, so
by transitivity and has the power of the continuum.
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for all real
,
.
Originally Posted by Chemboy
which proves that the norm is a continuous function. The proof is short, but is a bit of a trick. Take a look at 
and apply the triangle inequality.
...? I'm stuck. Very stuck. Please excuse any stupid things I say until I start to get this a little better.
...
like a 
and plug that 
back into the original, 
... Don't know if I'm allowed to do that though, or if it even gets me anywhere.
shouldn't it be
?
and complete the proof.
is defined by
.
unless 
, in which case 
.
such that 
. That may very well be completely pointless, that's just what I came up with while thinking about it. Please help me...I really really want to become good at this.
, as suggested by MagiMaster. Let me know if this works... (and please cut me some slack if I did anything stupid, I'm new at this. the phrase is "constructive criticism.")
: 
, so for 
: Since 
, we have 
and 
, so 
so we have 
so we have 
so 
to an infinite set 
. At least, that's what I think it means.
and an infinite amount of members in set 
the natural numbers, 
the rationals.
has the same cardinality as 
.
has the same cardinality as 
be a bijection 
. Then we can construct a bijection between 
and 
and letting it equal 
is surjective but definitely not bijective. Now every infinite set has a non-empty countable subset, say 
, so we have a bijection given by the restriction 
. This an isomorphism on sets, so that 
, since isomorphism is an equivalence relation.
i.e. they are disjoint. We also have that 
. Now the relation 
is symmetric, so 
. then by transitivity, 
.
. By reflexivity we have that 
. so that 
. Thus 
" or something of the sort. In general can we prove that it's true because 
?
. But the proof is trivial. Suppose that n>1, the multiplying both sides by n you have 
QED.
"
with rational end points is countable.
. You can even topologize that set with the order topology if you want to. 
. If you look at expressions for real numbers in base 2, say on the unit interval, it includes all sequences of 0's and 1's. It does not matter whether or not there are a finite number of 0"s (or equivalently a sequence that is eventually a string of 1's). So the proof boils down to showing that in base 2 any real number in the unit interval can be expressed as a sequence of 0's ad 1's and that the representation is unique except for at most a countable number of cases. (0.11111111...... = 1.00000000....... for instance)
be a countable subset of 
. Then 
. Now use the equivalence of the sets 
." 
because I didn't see how it represented everything in the unit interval...I didn't feel right about just throwing decimal places in there...
where 
for a given real number to get the desired unique representation.
where 
. Then 
for all 
and so S must be uncountable. i.e. 
by 
Its easy to see that f must be an injection so 
since 
and then using transitivity to show that 
, since we're not dealing with decimals that way? Or would I still have to show that there's a unique base-2 representation of every non-negative real number?
). What you actually showed was that card (
(where 
is the power set of 
, which is clearly greater than 
, so a one-to-one correspondence cannot be established.
since this is merely a restatement of the theorem in different terminology.
. This is easy to see if you write out the powerset of a set with, say, three elements. The powerset will have 
elements, all sets themselves. But of course this doesn't prove it, just an example to see.
, is it safe to generalize and prove that in general 
? The only sticky spot I see in that is the case where 
is indeed 
...
if 
but that has little to do with the problem since the gist of the issue involves infinite cardinals. Aslo note that in the trivial case, the power set of the empty set is the emnpty set, so you must exclude it. 
if A is a set, right? Just in case you don't, by definition it's the set of all functions
?
