I wasn't sure if this proof was correct or not.

Proposition. If $(a_n) \to a$ and $(a_n) \to b$, then $a = b$.

Proof. Suppose $(a_n) \to a$ and $(a_n) \to b$.

Then, $\lim \limits_{n \to \infty}(a_n)=a$ and $\lim \limits_{n \to \infty}(a_n)=b$.

So for every $\epsilon > 0$ there exists a $N \in N$ such that $n>N$ implies $|a_n -a| < \frac{\epsilon}{2}$ and a $M \in N$ such that $n>M$ implies $|a_n-b| < \frac{\epsilon}{2}$.

So $|(a_n-a)+(a_n-b)| < \frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon$.

So $|2a_n - (a + b)| < \epsilon$.

So $\lim \limits_{n \to \infty} 2a_n = a + b$.

So $2\lim \limits_{n \to \infty} a_n = a + b$.

So $\lim \limits_{n \to \infty} a_n = \frac{a + b}{2}$.

Case 1: $\lim \limits_{n \to \infty} a_n = a$

If $\lim \limits_{n \to \infty} a_n = a$, then $a = \frac{a+b}{2}$.

So $2a = a + b$ and $a = b$.

Case 2: $\lim \limits_{n \to \infty} a_n = b$

If $\lim \limits_{n \to \infty} a_n = b$, then $b = \frac{a+b}{2}$.

So $2b = a + b$ and $a = b$.

Therefore $a = b$.

  • 1
    $\begingroup$ Since we don't yet know whether or not limits are unique, we can't say with certainty that $a = \frac{a+b}{2}$ or $b = \frac{a+b}{2}$. How do we know that there isn't a third case, that $\lim_{n\to\infty}a_n = c$ for some $ c \in \mathbb{R}$? $\endgroup$
    – Matt E.
    May 21 at 23:58
  • 2
    $\begingroup$ Does this answer your question? Proving that a convergent sequence has a unique limit $\endgroup$
    – Jean L.
    May 22 at 0:41

3 Answers 3


I suggest that you adapt your proof using the following trick at the beginning: $$|a-b| = |a-b+(a_n-a_n)|= |(a-a_n)+(a_n-b)| \leq |a-a_n|+|b-a_n|$$ In particular you will get for all $n>M$ that $|a-b|\leq \epsilon$ which implies $a=b$, since $\epsilon>0$ is arbitrarily small.


Here is a better proof: If $a \neq b$. Assume $a < b$. Let $\epsilon = b - a$, then there is $N_1$ such that $n \ge N_1 \implies |a_n-a| < \dfrac{\epsilon}{2}$, and also $N_2$ such that $n \ge N_2 \implies |a_n - b| < \dfrac{\epsilon}{2}$. So if $n \ge N_1+N_2$, then $\epsilon \le |a_n-a|+|a_n-b| < \dfrac{\epsilon}{2}+\dfrac{\epsilon}{2} = \epsilon$, contradiction. So $a = b$.



Let $(X,d_{X})$ be a metric space.

If $(x_{n})_{n\in\mathbb{N}}$ is a convergent sequence of points in $X$, then its limit is unique.


We are going to prove the desired claim by contradiction.

Let $(x_{n})_{n\in\mathbb{N}}$ be a sequence of points in $X$ which converge to $a$ and $b$, respectively, where $a\neq b$.

On the one hand, the first convergence means that \begin{align*} (\forall\varepsilon\in\mathbb{R}_{>0})(\exists n_{1}\in\mathbb{N})(\forall n\in\mathbb{N})(n\geq n_{1} \Rightarrow d_{X}(x_{n},a) < \varepsilon) \end{align*}

On the other hand, the second convergence means that \begin{align*} (\forall\varepsilon\in\mathbb{R}_{>0})(\exists n_{2}\in\mathbb{N})(\forall n\in\mathbb{N})(n\geq n_{2} \Rightarrow d_{X}(x_{n},b) < \varepsilon) \end{align*}

Since $a\neq b$, we can take $3\varepsilon = d_{X}(a,b)$.

Consequently, there is $n_{\varepsilon} = \max\{n_{1},n_{2}\}$ such that for every $n\in\mathbb{N}$ satisfying $n\geq n_{\varepsilon}$ results that: \begin{align*} d_{X}(a,b) \leq d_{X}(x_{n},a) + d_{X}(x_{n},b) < 2\varepsilon = \frac{2d_{X}(a,b)}{3} \Rightarrow d_{X}(a,b) < 0, \end{align*} which is clearly a contradiction, because the metric is always non-negative.

Hopefully this helps!



Based on the suggestion of @MarekKryspin, consider a Hausdorff topological space $(X,\tau)$.

If $(x_{n})_{n\in\mathbb{N}}$ is a sequence of points in $X$ which converges, then the limit is unique.


Suppose the sequence $(x_{n})_{n\in\mathbb{N}}$ of points in $X$ converges to $a$ and $b$, respectively, where $a\neq b$.

Due to the definition of limits in topological spaces, the first convergence means that \begin{align*} (\forall N(a))(\exists n_{1}\in\mathbb{N})(\forall n\in\mathbb{N})(n\geq n_{1} \Rightarrow x_{n}\in N(a)). \end{align*}

Based on the same definition, the second convergence means that\begin{align*} (\forall N(b))(\exists n_{2}\in\mathbb{N})(\forall n\in\mathbb{N})(n\geq n_{2} \Rightarrow x_{n}\in N(b)). \end{align*}

Once the $(X,\tau)$ is Hausdorff, we can choose neighborhoods $N(a)$ and $N(b)$ s.t. $N(a)\cap N(b) = \varnothing$.

But this leads to a contradiction, because $x_{n}\in N(a)\cap N(b)$ whenever $n\geq\max\{n_{1},n_{2}\}$.

Hopefully this helps!

  • $\begingroup$ I was hoping for a topological proof of limit unicknes in Hausdorff spaces :) $\endgroup$ May 21 at 21:33
  • 1
    $\begingroup$ @MarekKryspin consider it done :-) $\endgroup$ May 21 at 21:48
  • $\begingroup$ hahahaha nice +1. $\endgroup$ May 21 at 21:51

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