If in a metric space a sequence converges to both $x$ and $y$, prove $x = y$

Question

I just started reading about metric spaces and am finding it hard to use all the definitions to come up with a proof. Would appreciate some help!

Answer 1

Hint: Use the triangular inequality. $$ d(x,y) \leq d(x,x_N) + d(x_N,y) < \epsilon, $$ where $N$ is sufficiently large such that $d(x,x_N)<\epsilon/2$ and $d(y,x_N)<\epsilon/2$.

Now since $\epsilon$ is arbitrarily small, $d(x,y)=0$ which implies $x=y$.

Answer 2

If $d(x_n,x) \to 0$ and $d(x_n,y) \to 0$ , then (triangle inequality !):

$0 \le d(x,y) \le d(x, x_n)+ d(x_n,y)$ for all $n$.

With $ n \to \infty$ we get $d(x,y)=0$, hence $x=y$.

Answer 3

The proof is identical to the following proof Just replace x and y by l and m

Her you are

Suppose that $L \neq M$. Let $\epsilon = |L - M|/2 > 0$. By hypothesis exists $N_1 \in \mathbb{N}$ such that $$ |a_n - L| < \dfrac{|L - M|}{2} \quad \text{if} \quad n \geq N_1 $$ By hypothesis, exists $N_2 \in \mathbb{N}$ such that $$ |a_n - M| < \dfrac{|L - M|}{2} \quad \text{if} \quad n \geq N_2 $$ Let $N = \max\{N_1,N_2\}$. If $n \geq N$, then by the triangle inequality $$ |L - M| = |(a_n - L) - (a_n - M)| < |a_ n - L| + |a_n - M| < 2\cdot \dfrac{|L - M|}{2} = |L - M| $$ Contradition!