# The Uniqueness of Limits of Functions on Metric Spaces

Recall from the Limits of Functions on Metric Spaces page that if $(S, d_S)$ and $(T, d_T)$ are metric spaces, $A \subseteq S$, $p \in S$ is an accumulation point of $A$, and $b \in T$, then we say that the limit of $f$ as $x$ approaches $p$ is equal to $b$ denoted $\displaystyle{\lim_{x \to p} f(x) = b}$ if for all $\epsilon > 0$ there exists a $\delta > 0$ such that if $x \in D(f) \setminus \{ p \}$ and $d_S(x, p) < \delta$ then:

(1)An equivalent definition is that $\displaystyle{\lim_{x \to p} f(x) = b}$ if for all $\epsilon > 0$ there exists a $\delta > 0$ such that if $x \in [D(f) \setminus \{ p \}] \cap B_S(p, \delta)$ then $f(x) \in B_T(b, \epsilon)$.

We will now look at a familiar theorem (familiar in the sense that we have seen a similar type of theorem a few times in the past!) which shows that limits of functions on metric spaces are unique.

Theorem 1: Let $(S, d_S)$ and $(T, d_T)$ be metric spaces, $A \subseteq S$, $p \in S$ be an accumulation point of $A$. If the limit of $f$ as $x$ approaches $p$ exists then it is unique, that is, for $b_1, b_2 \in T$, if $\displaystyle{\lim_{x \to p} f(x) = b_1}$ and $\displaystyle{\lim_{x \to p} f(x) = b_2}$ then $b_1 = b_2$. |

**Proof:**Let $\epsilon > 0$ be given.

- Since $\displaystyle{\lim_{x \to p} f(x) = b_1}$ we have that for $\epsilon_1 = \frac{\epsilon}{2} > 0$ that there exists a $\delta_1 > 0$ such that if $x \in D(f) \setminus \{ p \}$ and $d_S(x, p) < \delta_1$ then:

- Similarly, since $\displaystyle{\lim_{x \to p} f(x) = b_2}$ we have that for $\epsilon_2 = \frac{\epsilon}{2} > 0$ that there exists a $\delta_2 > 0$ such that if $x \in D(f) \setminus \{ p \}$ and $d_S(x, p) < \delta_2$ then:

- Now note that for $\delta = \min \{ \delta_1, \delta_2 \}$ we have that for all $x \in D(f) \setminus \{ p \}$ and $d_S(x, p) < \delta$ then both $(*)$ and $(**)$ holds simultaneously, and by the triangle inequality we have that:

- So for every $\epsilon > 0$ we have that $d_T(b_1, b_2) < \epsilon$ which implies that $d_T(b_1, b_2) = 0$. But this happens if and only if $b_1 = b_2$ which completes our proof. $\blacksquare$