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Consider the following statement:

Let $a,b\in\mathbb{R}$ be any two real numbers. Then we have that $$\forall\varepsilon>0:|a-b|<\varepsilon\implies a=b.$$

Here is my attempt to prove it:

Proof. Assume that $a\neq b$. Without loss of generality we may assume that $a<b$ (else we change the names of $a$ and $b$). $|a-b|=|b-a|=b-a>0$. Let $\varepsilon:=\frac{b-a}{2}>0$. By assumption we have that $$b-a=|a-b|<\varepsilon=\frac{b-a}{2}\implies 1<\frac{1}{2}$$ which is a contradiction. Therefore we must have that $a=b$.

q.e.d.

Is my proof correct? Can I use this statement to prove that the limit of a sequence $(a_n)_n$ is unique via assuming it has two limits $a,b$ and then concluding that for any $\varepsilon>0$ $$|a-b|\leq|a-a_n|+|a_n-b|<\varepsilon$$ for $n$ suffeciently large? (Because the statement then states that $a=b\implies$ uniqueness of limit).

Thank You

EDIT: I corrected the last line by replacing $=$ by $\leq$ since I used the triangle inequality.

mahlovic
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3 Answers3

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Yes, your answer is correct. Only one little mistake. Make sure that you use the triangle inequality in your last equation, $$ |a-b| \leq |a- a_n| + |a_n - b| < 2 \epsilon $$ and use the definition of sequence convergence: For any $\epsilon > 0$ there is a $N \in \mathbb{N}$, such that for all $n \geq N$ one has $$ |a- a_n| < \epsilon $$

Jannis H.
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In fact, according to your reasoning, if we suppose WLOG that $b > a$, we can also proceed as follows:

\begin{align*} \varepsilon = k(b-a) & \Rightarrow |a - b| = b - a < k(b-a)\\\\ & \Rightarrow (1 - k)(b - a) < 0\\\\ & \Rightarrow b - a < 0\\\\ & \Rightarrow b < a \end{align*}

whence we get a contradiction (if we assume that $k\in(0,1)$).

If you want to prove that limits are unique, you can proceed more generally as it follows.

Suppose we are given a metric space $(X,d_{X})$ and a sequence $x_{n}\in X$ which converges to $L$.

Let us also assume that $x_{n}$ converges to $M$. We shall prove that $L = M$.

Indeed, if $L\neq M$, according to the definition of convergence in metric spaces, precisely: \begin{align*} (\forall\varepsilon > 0)(\exists n_{\varepsilon}\in\mathbb{N})(\forall n\in\mathbb{N})(n\geq n_{\varepsilon}\Rightarrow d_{X}(x_{n},L) < \varepsilon) \end{align*}

We can take $\varepsilon = kd_{X}(L,M)$, where $k\in(0,1)$.

Consequently, there are $n^{1}_{\varepsilon}\in\mathbb{N}$ and $n^{2}_{\varepsilon}\in\mathbb{N}$ such that

\begin{align*} \begin{cases} n\geq n^{1}_{\varepsilon} \Rightarrow d_{X}(x_{n},L) < \varepsilon/2\\\\ n\geq n^{2}_{\varepsilon} \Rightarrow d_{X}(x_{n},M) < \varepsilon/2 \end{cases} \end{align*}

we take $n_{\varepsilon} = \max\{n^{1}_{\varepsilon},n^{2}_{\varepsilon}\}$, we conclude that \begin{align*} n\geq n_{\varepsilon} & \Rightarrow d_{X}(L,M) \leq d_{X}(x_{n},L) + d_{X}(x_{n},M) \leq \varepsilon = kd_{X}(L,M) \end{align*}

whence we conclude that $L = M$, a contradiction.

Hopefully this helps!

Átila Correia
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Your answer is fine, but there is no reason to consider the two cases $a<b$ and $a>b,$ even if you only eliminate one or the using a WLOG argument.

Lemma: For real $x\geq 0$ and $x<\varepsilon,$ for all $\varepsilon>0,$ then $x=0.$

Proof: Esssentially the same. If $x\neq 0,$ then $x>0,$ and we can let $\varepsilon =x/2>0.$

Then your result follows with $x=|a-b|,$ since it is easy to show $|a-b|=0$ iff $a=b.$

This argument works with $a,b$ complex numbers, too, since $|a-b|\geq 0,$ even though we don’t have a notion of order in complex numbers.

It also works in any metric space, with $|a-b|$ replaced by the metric $d(a,b)$ of the metric space.

Thomas Andrews
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