Question

Is the proof for the following proposition correct? Can it be simplified?

Answer 1

Proposition:
Let \(x_1, x_2, ..., x_n, ...\) be a convergent sequence of points in a metric space \((X, d)\). Further, let \(x\) and \(y\) be points in \((X, d)\) such that \(xn \to x\) and \(xn \to y\). Then \(x = y\).

Proof:
Assume that \(x \ne y\) then by the definition of a metric \(d(x,y) > 0\). Set \(\epsilon = \frac{1}{3} d(x,y)\) and choose some \(n > N\) where \(d(x,x_n\) and \(d(y,x_n\) less than \(\epsilon\). This is possibile by the definition of a convergent sequence. By the triangle equality:
\[d(x,y) \le d(x,x_n) + d(y,x_n) \lt 2 * \epsilon = \frac{2}{3}d(x,y)\]
This is a contradiction and therefore \(d(x,y)=0\) which implies \(x=y\).

Answer 2

Now without typos (@pre-algebra: wanna give it a try? ;-) )

Proposition:
Let \(x_1, x_2, ..., x_n, ...\) be a convergent sequence of points in a metric space \((X, d)\). Further, let \(x\) and \(y\) be points in \((X, d)\) such that \(x_n \to x\) and \(x_n \to y\). Then \(x = y\).

Proof:
Assume that \(x \ne y\) then by the definition of a metric \(d(x,y) > 0\). Set \(\epsilon = \frac{1}{3} d(x,y)\) and choose some \(n > N\) where \(d(x,x_n)\) and \(d(y,x_n)\) less than \(\epsilon\). This is possibile by the definition of a convergent sequence. By the triangle equality:
\[d(x,y) \le d(x,x_n) + d(y,x_n) \lt 2 * \epsilon = \frac{2}{3}d(x,y)\]
This is a contradiction and therefore \(d(x,y)=0\) which implies \(x=y\).

Answer 3

@mwande03 , @satellite73 : maybe one of you?

Answer 4

@pre-algebra : mentioning you didn't work the first time, so trying again...

Answer 5

let me help you out jerico:
@across @JamesJ @Zarkon help with proof?

Answer 6

Thanks @TuringTest ...