Exercise 1.1.14 (Uniqueness of limits)

Consider the metric space $(X,d)$ and let ${(x^{(n)})}_{n=m}^{\infty }$ be a sequence in $X$ with two limits, $x$ and $x^{\prime }$. The limit of a sequence in a metric space is unique, i.e., $x=x^{\prime }$.

To prove this, let us note that $\forall <!--\; FUNCTION\; APPLICATION\; -->(𝜖>0)\exists <!--\; FUNCTION\; APPLICATION\; -->(N\ge m)\forall <!--\; FUNCTION\; APPLICATION\; -->(n\ge N)d(x^{(n)},x)<𝜖$, i.e., our sequence converges to $x$. Similarly, $\forall <!--\; FUNCTION\; APPLICATION\; -->({𝜖}^{\prime }>0)\exists <!--\; FUNCTION\; APPLICATION\; -->(N^{\prime }\ge m)\forall <!--\; FUNCTION\; APPLICATION\; -->(n\ge N^{\prime })d(x^{(n)},x)<{𝜖}^{\prime }$. If we want to show that $x=x^{\prime }$, but we only have access to information based on limiting processes, we can use the fact that $\forall <!--\; FUNCTION\; APPLICATION\; -->(𝜖>0)x-x^{\prime }<𝜖⟺x=x^{\prime }$ to try to constrain the difference between $x$ and $x^{\prime }$ under some $\delta$ which can be chosen arbitrarily above zero.

Let us consider some arbitrary $\delta >0$. Choose some $𝜖,{𝜖}^{\prime }<\delta ∕2$ such that $d(x,x^{\prime })\le d(x,x^{(n)})+d(x^{(n)},x^{\prime })<\delta ∕2+\delta ∕2=\delta$. By a previous statement given above, this means that $x=x^{\prime }$.