Problem: let [imath]E \subseteq \mathbb{R}^n[/imath] and let [imath]\mathbf{y}[/imath] be a limit point of [imath]E[/imath]. Prove that there exists a sequence [imath]\mathbf{x}^h \ne \mathbf{y}[/imath] of elements of [imath]E[/imath] such that [imath]\mathbf{x}^h \to \mathbf{y}[/imath] when [imath]h \to +\infty[/imath].
In the following work, with [imath]B_r(\mathbf{a})[/imath] I mean the ball of radius [imath]r > 0[/imath] and center [imath]\mathbf{a} \in \mathbb{R}^n[/imath].
My work: by hypothesis [imath]\mathbf{y}[/imath] be a limit point of [imath]E[/imath], hence for each [imath]\epsilon > 0[/imath] we have [imath](B_\epsilon(\mathbf{y}) \setminus \{\mathbf{y}\}) \cap E \ne \varnothing[/imath]. Being the intersection non-empty, there exists [imath]\mathbf{x}^\epsilon \in (B_\epsilon(\mathbf{y}) \setminus \{\mathbf{y}\}) \cap E[/imath]. By definition of intersection, this means that [imath]\mathbf{x}^\epsilon \ne \mathbf{y}[/imath] and [imath]\mathbf{x}^\epsilon \in E[/imath]. Let [imath]h \in \mathbb{N}[/imath], choosing [imath]\epsilon = 1/(h+1)[/imath], from [imath]\mathbf{x}^{1/(h+1)} \in B_{1/(h+1)}(\mathbf{y})[/imath] it follows that [imath]\mathbf{x}^{1/(h+1)} \to \mathbf{y}[/imath] when [imath]h \to +\infty[/imath] because, from the definition of ball in [imath]\mathbb{R}^n[/imath], we have [imath]\| \mathbf{x}^{1/(h+1)}-\mathbf{y} \| < \frac{1}{h+1} \to 0[/imath] when [imath]h \to +\infty[/imath].
Is my proof correct?
In the following work, with [imath]B_r(\mathbf{a})[/imath] I mean the ball of radius [imath]r > 0[/imath] and center [imath]\mathbf{a} \in \mathbb{R}^n[/imath].
My work: by hypothesis [imath]\mathbf{y}[/imath] be a limit point of [imath]E[/imath], hence for each [imath]\epsilon > 0[/imath] we have [imath](B_\epsilon(\mathbf{y}) \setminus \{\mathbf{y}\}) \cap E \ne \varnothing[/imath]. Being the intersection non-empty, there exists [imath]\mathbf{x}^\epsilon \in (B_\epsilon(\mathbf{y}) \setminus \{\mathbf{y}\}) \cap E[/imath]. By definition of intersection, this means that [imath]\mathbf{x}^\epsilon \ne \mathbf{y}[/imath] and [imath]\mathbf{x}^\epsilon \in E[/imath]. Let [imath]h \in \mathbb{N}[/imath], choosing [imath]\epsilon = 1/(h+1)[/imath], from [imath]\mathbf{x}^{1/(h+1)} \in B_{1/(h+1)}(\mathbf{y})[/imath] it follows that [imath]\mathbf{x}^{1/(h+1)} \to \mathbf{y}[/imath] when [imath]h \to +\infty[/imath] because, from the definition of ball in [imath]\mathbb{R}^n[/imath], we have [imath]\| \mathbf{x}^{1/(h+1)}-\mathbf{y} \| < \frac{1}{h+1} \to 0[/imath] when [imath]h \to +\infty[/imath].
Is my proof correct?
