Sequence ........ The sequence $ \left \{x_n \right \} \subset \mathbb {R} $ has a finite limit $ a \in \mathbb {R} $. Prove that
$$\bigcap\limits_{\alpha >0}\bigcup\limits_{\beta >0}\bigcap\limits_{n>\beta }\left ( x_n-\alpha ,x_n+\alpha  \right )=\left \{ a \right \}$$
I tried to solve through limit definition
$$\lim_{n\rightarrow \infty }x_n=a\Leftrightarrow \forall \alpha >0 \; \exists \beta >0 \; \forall n>\beta  \; \left | a-x_n \right |<\alpha \Leftrightarrow$$
$$\Leftrightarrow \forall \alpha >0 \; \exists \beta >0 \; \forall n>\beta \; a\in \left ( x_n-\alpha ,x_n+\alpha  \right )\Leftrightarrow $$
$$\Leftrightarrow \forall \alpha >0 \; \exists \beta >0 \; a\in \bigcap\limits_{n>\beta }\left ( x_n-\alpha ,x_n+\alpha  \right ) $$
The question is whether it is necessary to prove that there is only one limit?
If so, how to prove it?

I was able to prove the uniqueness of the limit!
$$\exists \lim_{n\rightarrow \infty }x_n=B\neq A\Leftrightarrow \forall \varepsilon  >0 \; \exists \delta \left ( \varepsilon  \right )>0\big|\forall n>\delta  \; x_n\in U_{\varepsilon }\left ( B \right )$$
\begin{multline*}
\varepsilon \leq \frac{\left | B-A \right |}{2}\Rightarrow U_{\varepsilon}\left ( B \right )\cap U_{\varepsilon }\left ( B \right )=\varnothing \Rightarrow B=A\Leftrightarrow \exists !\lim_{n\rightarrow \infty }x_n=A\Rightarrow \\\\ \Rightarrow \bigcap\limits_{\varepsilon  >0}\bigcup\limits_{\delta  >0}\bigcap\limits_{n>\delta  }U_{\varepsilon }U_{\varepsilon }\left ( x_n \right )=\left \{ A \right \}\end{multline*}
 A: I think what you have done is sufficient, but I will express it a little differently.
$$(1)\quad\lim_{n\rightarrow \infty }x_n=t\Leftrightarrow \forall \alpha >0 \; \exists \beta >0 \; \forall n>\beta,  \; \left | t-x_n \right |<\alpha \Leftrightarrow$$
$$\Leftrightarrow \forall \alpha >0 \; \exists \beta >0 \; \forall n>\beta, \; t\in \left ( x_n-\alpha ,x_n+\alpha  \right )\Leftrightarrow $$
$$\Leftrightarrow \forall \alpha >0 \; \exists \beta >0, \; t\in \bigcap\limits_{n>\beta }\left ( x_n-\alpha ,x_n+\alpha  \right ).$$
Since $\lim_{n\rightarrow \infty }x_n=a$ is given, the above shows: $\left \{ a \right \}\subset\bigcap\limits_{\alpha >0}\bigcup\limits_{\beta >0}\bigcap\limits_{n>\beta }\left ( x_n-\alpha ,x_n+\alpha  \right ).$
We still must show the reverse inclusion. It is enough to show that whenever $b \neq a$, that $b$ does not belong to $ \bigcap\limits_{\alpha >0}\bigcup\limits_{\beta >0}\bigcap\limits_{n>\beta }\left ( x_n-\alpha ,x_n+\alpha  \right ).$
But this follows directly from (1), since a limit is unique. I don't think you should prove the general fact that a limit is unique in the context of this problem. The proof is not hard, and exists on math stack exchange, here: https://math.stackexchange.com/a/882262/688046
