極限の一意性

証明

$ \forall a_\bullet:\N\to X\forall\alpha,\beta\in X:

$ \begin{dcases}a_n\to\alpha\quad(n\to\infty)\\a_n\to\beta\quad(n\to\infty)\end{dcases}

$ \iff\forall\varepsilon>0:\begin{dcases}\exist N\in\N\forall n>N:d(a_n,\alpha)<\varepsilon\\\exist N\in\N\forall n>N:d(a_n,\beta)<\varepsilon\end{dcases}

$ \implies\forall\varepsilon>0\exist N_1,N_2\in\N\forall n>\max\{N_1,N_2\}:\begin{dcases}d(a_n,\alpha)<\varepsilon\\d(a_n,\beta)<\varepsilon\end{dcases}

$ \implies\forall\varepsilon>0\exist N_1,N_2\in\N\forall n>\max\{N_1,N_2\}:d(\alpha,\beta)\le d(\alpha,a_n)+d(a_n,\beta)<\varepsilon

$ \implies\forall\varepsilon>0:d(\alpha,\beta)<2\varepsilon

$ \implies(

$ d(\alpha,\beta)\neq0

$ \implies d(\alpha,\beta)<2\cdot\frac14d(\alpha,\beta)

$ \iff\frac12d(\alpha,\beta)<0

$ \iff\bot

$ )

$ \implies d(\alpha,\beta)=0

$ \iff\alpha=\beta

$ \underline{\therefore\forall a_\bullet:\N\to X\forall\alpha,\beta\in X:\begin{dcases}a_n\to\alpha\quad(n\to\infty)\\a_n\to\beta\quad(n\to\infty)\end{dcases}\implies\alpha=\beta\quad}_\blacksquare