# How do we prove the limit of a sequence of real numbers is unique?

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 $$a = b$$.

• 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 $c \in \mathbb{R}$? May 21 at 23:58
• Does this answer your question? Proving that a convergent sequence has a unique limit May 22 at 0:41

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 $$a = b$$.

Claim

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.

Proof

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!

EDIT

Claim

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.

Proof

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!

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