N is an infinite subset of R with no cluster point

[G-X]

[Math]Definitions:[/Math] Let [Math]\epsilon > 0[/Math]. A neighborhood (or [Math]\epsilon[/Math]-neighborhood) of a point [Math]a[/Math] is the set [Math]N(a, \epsilon) =[/Math] [Math]\{x \in R : d(a, x) < \epsilon\}[/Math]. A point [Math]a[/Math] is an accumulation point of a set E if every neighborhood of [Math]a[/Math] contains a point [Math]x \neq a[/Math] such that [Math]x \in E[/Math]. That is, [Math]a[/Math] is an accumulation point of the set E if [Math]E \cap (N(a, \epsilon) / \{a\}) \neq \varnothing[/Math] for all [Math]\epsilon > 0[/Math].

We know [Math]N \subset R^{k}[/Math]with [Math]k = 1[/Math] or [Math]2[/Math]. By definition, [Math]N[/Math], is an infinite subset of [Math]R^{k}[/Math].

We will prove that there exists an infinite subset of [Math]R^{k}[/Math] that has no cluster points. Let [Math]a \in N[/Math] and [Math]0 < r < \dfrac{1}{2}[/Math]. We can define an open ball [Math]B_{r}(a) =[/Math] [Math]\{y \in R^{k}: d(y, a) < r\}[/Math]. Let [Math]d = \dfrac{1}{2}[/Math]. Let [Math]x \in N[/Math] and [Math]x \neq a[/Math]. We can conclude that [Math]d(y, a) < r = d < d(x, a)[/Math]. This means that [Math]E \cap (B_{r}(a) / \{a\}) = \varnothing[/Math] and thus we have proven that there exists an infinite subset of [Math]R^{k}[/Math] with no cluster point.

 

[HallsofIvy]

I don't understand your proof.

[Math]Definitions:[/Math] Let [Math]\epsilon > 0[/Math]. A neighborhood (or [Math]\epsilon[/Math]-neighborhood) of a point [Math]a[/Math] is the set [Math]N(a, \epsilon) =[/Math] [Math]\{x \in R : d(a, x) < \epsilon\}[/Math]. A point [Math]a[/Math] is an accumulation point of a set E if every neighborhood of [Math]a[/Math] contains a point [Math]x \neq a[/Math] such that [Math]x \in E[/Math]. That is, [Math]a[/Math] is an accumulation point of the set E if [Math]E \cap (N(a, \epsilon) / \{a\}) \neq \varnothing[/Math] for all [Math]\epsilon > 0[/Math].

We know [Math]N \subset R^{k}[/Math]with [Math]k = 1[/Math] or [Math]2[/Math]. By definition, [Math]N[/Math], is an infinite subset of [Math]R^{k}[/Math].

What is "N"? You have not defined set "N".

We will prove that there exists an infinite subset of [Math]R^{k}[/Math] that has no cluster points. Let [Math]a \in N[/Math] and [Math]0 < r < \dfrac{1}{2}[/Math]. We can define an open ball [Math]B_{r}(a) =[/Math] [Math]\{y \in R^{k}: d(y, a) < r\}[/Math]. Let [Math]d = \dfrac{1}{2}[/Math]. Let [Math]x \in N[/Math] and [Math]x \neq a[/Math]. We can conclude that [Math]d(y, a) < r = d < d(x, a)[/Math]. This means that [Math]E \cap (B_{r}(a) / \{a\}) = \varnothing[/Math]

And what is "E"? You have not defined "E"!

and thus we have proven that there exists an infinite subset of [Math]R^{k}[/Math] with no cluster point.

 

[G-X]

Made slight adjustments - Changed E to N. N is just the natural numbers. Do I even need, d?

[Math]Definitions:[/Math] Let [Math]\epsilon > 0[/Math]. A neighborhood (or [Math]\epsilon[/Math]-neighborhood) of a point [Math]a[/Math] is the set [Math]N(a, \epsilon) =[/Math] [Math]\{x \in R : d(a, x) < \epsilon\}[/Math]. A point [Math]a[/Math] is an accumulation point of a set E if every neighborhood of [Math]a[/Math] contains a point [Math]x \neq a[/Math] such that [Math]x \in E[/Math]. That is, [Math]a[/Math] is an accumulation point of the set E if [Math]E \cap (N(a, \epsilon) / \{a\}) \neq \varnothing[/Math] for all [Math]\epsilon > 0[/Math].

We know [Math]N \subset R^{k}[/Math]with [Math]k = 1[/Math] or [Math]2[/Math]. By definition, [Math]N[/Math], is an infinite subset of [Math]R^{k}[/Math] as N is the natural numbers.

We will prove that there exists an infinite subset of [Math]R^{k}[/Math] that has no cluster points. Let [Math]a \in N[/Math] and [Math]0 < r < \dfrac{1}{2}[/Math]. We can define an open ball [Math]B_{r}(a) =[/Math] [Math]\{y \in R^{k}: d(y, a) < r\}[/Math]. Let [Math]d = \dfrac{1}{2}[/Math]. Let [Math]x \in N[/Math] and [Math]x \neq a[/Math]. We can conclude that [Math]d(y, a) < r < d < d(x, a)[/Math]. This means that [Math]N \cap (B_{r}(a) / \{a\}) = \varnothing[/Math] and thus we have proven that there exists an infinite subset of [Math]R^{k}[/Math] with no cluster point.