Add 2 Dense Sets for Non-Dense Set Result

[cragar]

I can add 2 dense sets together and get a non dense set right?

 

[HallsofIvy]

I take it you mean "take the union" rather than "adding" in the numerical sense. A set, A, is said to be "dense" in another set, B, if and only if every point of B is in some delta-neighborhood of a point in A. If that is true for two sets, say X and Y are both dense in Z then certainly their union is dense in Z.

 

[cragar]

A dense set means its uncountable right? I was reading in a book and it said take the set of all real numbers and we will call these the blue numbers, then take a second set of the real numbers and call them the red numbers . Now take the union of these sets and put each number after itself, red number then a blue number. Does this work. I think the book is called infinity.

 

[tiny-tim]

hi cragar!

A dense set means its uncountable right?

nope

eg the Cantor set is uncountable but nowhere dense …

see http://en.wikipedia.org/wiki/Cantor_set#Properties" ​

 

[cragar]

And uncountable meaning If I am at one number I couldn't tell you the next number in the list.

 

[tiny-tim]

And uncountable meaning If I am at one number I couldn't tell you the next number in the list.

depends what you mean by "list"

 

[cragar]

Like the set of real numbers is uncountable . And I thought that means that if I am at 0 there is no next number to the right of zero on the continuum? Or am I wrong .

 

[tiny-tim]

i can list the set of real numbers this way …

for any number x in [0,1), i make a sub-list {… x-3 x-2 x-1 x x+1 x+2 …}, and i order all the sub-lists in order of x …

that way, there is always a next number! ​

 

[micromass]

Like the set of real numbers is uncountable . And I thought that means that if I am at 0 there is no next number to the right of zero on the continuum? Or am I wrong .

No, that's not at all what it means! Uncountable simply means that there is no injection from the set to the natural numbers.
The real numbers indeed have the property that next to 0 there is no next number, but not every uncountable set has that property.

 

[cragar]

so uncountable means that I can't put the set into a one-to-one correspondence with the natural numbers. And are all uncountable sets infinite?
Thanks for taking the time to answer my questions

 

[micromass]

so uncountable means that I can't put the set into a one-to-one correspondence with the natural numbers.

Hmm, not exactly, because finite sets can also not be put in one-to-one correspondance with the naturals. But you could define uncountable as that the set is finite and you can't put it into a one-to-one correspondence with the naturals.

And are all uncountable sets infinite?

Yes!

 

[cragar]

If I excluded finite sets would that definition work . And I am still not sure what a dense set is.

 

[micromass]

If I excluded finite sets would that definition work.

Yes, then it would work.

And I am still not sure what a dense set is.

A set D is dense in X if $\overline{D}=X$. It has nothing to do with countability, but everything with topology and metric spaces.

A characterization for metric spaces that I find very helpful:
A set D is dense in X if for every x in X, there exists a sequence (x_n) in D such that $x_n\rightarrow x$.

 

[cragar]

Im not sure I understand you definition of dense . When you say a set D is dense in X, What is X is it a set , and then you say x_n goes to x , Are you saying I can match up these elements? Thanks for your taking the time to explain this to me.

 

[micromass]

Im not sure I understand you definition of dense . When you say a set D is dense in X, What is X is it a set,

X can be the underlying metric space, or topological space. And D is a subset of X. For example, we could have $X=\mathbb{R}$ and $D=\mathbb{R}\setminus \{0\}$.

and then you say x_n goes to x , Are you saying I can match up these elements?

What do you mean, "match up the elements"? I mean that x_n converges to x, that is

$$\forall \varepsilon >0:~\exists n_0:~\forall n\geq n_0:~d(x,x_n)<\varepsilon$$.

 

[cragar]

is x an element of D . and when you say $x_n$ converges to x , is this like a limit ?

 

[micromass]

is x an element of D .

No, x does not need to be an element of D.

and when you say $x_n$ converges to x , is this like a limit ?

Yes, this is like a limit.

I should have asked you a long time ago: but what math classes did you already take? And what made you ask this question. Maybe we can give you a better answer depending on that information...

 

[cragar]

All of my math background is applied math. calculus and differential equations. I am a physics major. I am taking discrete math this summer so maybe I should wait. My original question was can I have the union of 2 dense sets and get a non-dense set.

 

[micromass]

All of my math background is applied math. calculus and differential equations. I am a physics major. I am taking discrete math this summer so maybe I should wait. My original question was can I have the union of 2 dense sets and get a non-dense set.

OK, thanks. But what made you ask this question? Where did you encounter the notion of "dense set"?

And no, it isn't possible for the union of two dense sets to be non-dense...

 

[cragar]

I was reading a book called infinity and it talked about dense sets and it said you could have the union of 2 dense sets and get a non-dense set, but I guess the book could be wrong. Could I have the union of 2 uncountable sets and make it a countable set.

 

[micromass]

I was reading a book called infinity and it talked about dense sets.

Hmm, in that point-of-view, I'm doubting that the author of the book uses the same dense as we use. There are more definitions of "density" out there, so perhaps he's using another one.

Could I have the union of 2 uncountable sets and make it a countable set.

No, uncountable means that the set is big. And the union of two big sets is an even bigger set. Thus it remains uncountable.

 

[cragar]

Is the smallest infinity the set of natural numbers?

 

[micromass]

Is the smallest infinity the set of natural numbers?

Yes!

 

[cragar]

but aren't there an infinite number of positive even numbers which would be a subset of the naturals.

 

[micromass]

but aren't there an infinite number of positive even numbers which would be a subset of the naturals.

There are as many positive even numbers as there are natural numbers. Indeed, we have a one to one correspondence:

$$\mathbb{N}\rightarrow\{\text{even numbers}\}:n\rightarrow 2n$$

And if there is a one-to-one correspondence between two sets, then these sets have equal size. It might seems paradoxical that a proper subset has as many elements as the superset, but that's something we have to live with. It is a situation that arises whenever we deal with infinity.

 

[cragar]

How is infinity defined?

 

[micromass]

A set is infinite if it isn't finite. And a set X is finite, if it is empty or if there exists a one-on-one correspondence $X\rightarrow \{1,...,n\}$ for a natural number n.

 

[cragar]

ok and why can we say that there are more reals than naturals . I mean they are both infinite. I have seen cantors diagonal argument.

 

[micromass]

ok and why can we say that there are more reals than naturals . I mean they are both infinite. I have seen cantors diagonal argument.

There are different sizes of infinity. Two sets are said to have equal cardinality (=equal size) if there exists a one-on-one correspondence between them. A set A is said to have less or equal cardinality than B if there exists an injection $A\rightarrow B$. So a set A has strictly less cardinality if there exists an injection $A\rightarrow B$ but there does not exists a bijection $A\rightarrow B$.

So, it is very easy to see why the naturals have less cardinality than the reals. Indeed, consider

$$\mathbb{N}\rightarrow \mathbb{R}:~n\rightarrow n$$,

this is an injection. So the cardinality of N is less (or equal!) to the cardinality of R. But, in fact, the cardinality is striclty less. For that, we need to show that there does not exist a bijection between N and R, and this is what Cantor's diagonal argument shows.

 

[cragar]

Ok I see , I am very much enjoying this conversation .