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Can there be no interior point in ℕ?

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Depends on the topology. Are you assuming the one inherited from the standard on R?
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>I read somewhere that there exists no interior point in ℕ

That's true if you are thinking of ℕ as a subset of the reals with their usual topology, but...

> as every neighborhood of x (an arbitrary element belonging to ℕ) doesn't belong to ℕ.

this explanation for it is badly phrased at best. A neighborhood is not "an arbitrary element".

One way to define the boundary is this: a point p is called a "boundary point" of a set S if every neighborhood of p contains at least one point in S and one point not in S. Using this definition, it is easy to show that every point in ℕ is a boundary point, ie, not an interior point.

>I mean how could this be? Isn't 2 and 4 neighborhood of 3?

Not in ℝ, no. The interval [2,4] is a neighborhood of 3, but {2,4} is not.
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In a topology (X,T), a subset S⊆X has no interior iff there are no open sets contained in S (other than ∅).

In ℝ with the usual topology, no nonempty open sets are subsets of ℕ. There are other topologies on ℝ under which ℕ would have nonempty interior, such as the topology generated by {{a+bk | k∈ℕ} | a,b∈ℝ}.

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