Is the Natural Numbers Dense in Itself?

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SUMMARY

The discussion confirms that the natural numbers (\mathbb{N}) are dense in themselves when considering the topological definition of denseness, where every open set containing a point in \mathbb{N} must also contain another point from \mathbb{N}. However, it clarifies that there are various definitions of denseness, and under certain definitions, \mathbb{N} is not dense in itself. The conversation emphasizes the importance of specifying the definition of denseness being used, as it can lead to confusion, particularly when comparing \mathbb{N} with other sets like the real numbers (\mathbb{R}).

PREREQUISITES
  • Understanding of topological spaces
  • Familiarity with the concept of denseness in mathematics
  • Knowledge of different definitions of denseness
  • Basic comprehension of order-theoretic concepts
NEXT STEPS
  • Research the various definitions of denseness in topology
  • Explore the concept of dense-in-itself sets
  • Learn about the differences between denseness in \mathbb{N} and \mathbb{R}
  • Investigate order-theoretic notions of denseness
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Mathematicians, students of topology, and anyone interested in the properties of number sets and their relationships in different mathematical contexts.

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\ Is \ \mathbb{N} \ dense \ in \ itself.
 
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Yes, every space is dense in itself.
 
"A is dense in B" (with A and B topological spaces) mean "given any point p in B, every open set containing p contains some point of A." Of course, if A= B, that is trivially true.
 
HallsofIvy said:
"A is dense in B" (with A and B topological spaces) mean "given any point p in B, every open set containing p contains some point of A." Of course, if A= B, that is trivially true.

So is there a point p\in(n,n+1)\forall n\in\mathbb{N} such that p\in\mathbb{N}?
 
mjpam said:
So is there a point p\in(n,n+1)\forall n\in\mathbb{N} such that p\in\mathbb{N}?

No, but that doesn't matter. We're talking about denseness of N in N. Your example doesn't apply because you're confused with showing that N is dense in R!Also, for the OP, note that there are different (non-equivalent) definitions of denseness. Most often dense is applied in topological spaces, and this is what people in this thread do. But there are other definitions of denseness such that N is not dense in N. I'm just saying this because this is probably what confuses you. But you should always check what definition of denseness you are using.
 
micromass said:
No, but that doesn't matter. We're talking about denseness of N in N. Your example doesn't apply because you're confused with showing that N is dense in R!Also, for the OP, note that there are different (non-equivalent) definitions of denseness. Most often dense is applied in topological spaces, and this is what people in this thread do. But there are other definitions of denseness such that N is not dense in N. I'm just saying this because this is probably what confuses you. But you should always check what definition of denseness you are using.

No, I just misread the original statement to say that "every open subset of A must contains a member of A". After re-reading the original statement I see how, by that definition of "dense", every set is trivially dense in itself.
 
micromass said:
Also, for the OP, note that there are different (non-equivalent) definitions of denseness. Most often dense is applied in topological spaces, and this is what people in this thread do. But there are other definitions of denseness such that N is not dense in N. I'm just saying this because this is probably what confuses you. But you should always check what definition of denseness you are using.
What kind of 'denseness' do you have in mind here? Some measure theoretic concept?
 
No, there are some order-theoretic notions of denseness. For example:

for every x and y, there exists a z such that x<z<y. This is sometimes called denseness. I just mention this, because the OP has talked about this in another post. I just wanted to clear up what definition of denseness we're using here...
 
Right. So the order on N is definitely not dense.

And then there's the term dense-in-itself, meaning 'containing no isolated points' (which of course depends on the superset you're considering.)
 

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