- #1

Eclair_de_XII

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- 91

## Homework Statement

"Let ##E \subset ℝ##. Prove that ##E## is closed if for each ##x_0##, there exists a sequence of ##x_n \in E## that converges to ##x_0##, it is true that ##x_0\in E##. In other words, prove that ##E## is closed if it contains every limit of sequences for each of its elements."

**2. Homework Equations**

__Closed:__A set is closed if it contains all of its accumulation points.

## The Attempt at a Solution

Choose an arbitrary ##x \in E##. Then there exists a sequence ##\{x_n\}_{n=1}^\infty## that converges to ##x##, where ##x_n\in E,\forall n\in ℕ##. Let ##\epsilon>0##. Then there exists an ##N\in ℕ## such that if ##n\geq N##, then ##|x_n - x|<\epsilon##. Equivalently, for ##n\geq N##, ##x_n\in (x-\epsilon,x+\epsilon)##. This neighborhood of ##x## contains all but finitely many ##x_n\in E##. Moreover, any neighborhood of ##x## contains infinitely many ##x_n \in E##. Therefore, ##x## is an accumulation point of ##E##. Since ##x\in E##, then it is implied that any ##x \in E## is an accumulation point of ##E##. The set ##E## is closed, as a result, since it contains all of its accumulation points.

I felt like this proof was too easy. Is there something I'm missing?