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I am trying to show that the interval $[0,1)$ is a closed subset of $(-1,1)$ by using the definition that a closed subset contains all of its limit points. So for a convergent sequence $\{x_n\}$ in $[0,1)$ we have that $0 \leq x_{n} < 1$ for all $n \in \mathbf{N}$. How can I show that $\lim_{n \rightarrow \infty}x_{n} = x$ implies that $x \in [0,1)$? I understand that 1 cannot be a limit point of $[0,1)$ since $1 \not\in (-1,1)$ but I'm having a hard time saying that in a formal way?

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  • $\begingroup$ Note that any element of $(-1,1)$ must satisfy $-1 < x <1$. Hence if $x_n \to x$ then you must have $-1 < x < 1$. $\endgroup$ – copper.hat Sep 18 '16 at 17:24
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To show that $1$ is not in the adherence, observe that $$(1-\varepsilon,1)\cap (-1,1)=\emptyset.$$ In other words, there is no sequence of $[0,1)$ that converge to $1$.

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I think you missed an important part of the definition which made the problem harder. The statement you want to prove is for any sequence $\{x_n\}$ in $[0,1)$ such that $$\lim_n x_n = x \quad \text{ for some } x\in (-1,1),$$ then we must have $x\in [0,1)$.

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