$$B_{\epsilon}(x)=\cases{\{x\} & $0<\varepsilon \le 1$\\
X & $\varepsilon > 1$}$$
Open Ball in Standard Discrete Metric Space
Is this proof wrong?
I have no trouble understanding the first half of the proof, which for all $x$ and $a$ with a distance greater than $1$ is not in the open ball with a radius less than $1$ and such open ball contains the centre element itself. However, in the second half, just so wired that $$\forall x\in A:d(x, a)>\epsilon\implies x\in B_{\epsilon}(a;d)$$ How does two elements distance greater than $\epsilon$ and one element will fall in the open ball that centre at the other elements with a radius less than the distance? Is this constructive approach wrong?
I know what the discrete metric is
Appreciate any comments
Best Answer
You're being overly formal IMO.
Let $0 < \varepsilon \le 1$.
Then $y \in B_\varepsilon(x)$ iff $d(x,y) < \varepsilon \le 1$ so $d(x,y) <1$ and as the only values of $d$ are $0$ and $1$ it follows that $d(x,y)=0$ so that $y=x$. So $y \in B_\varepsilon(x) \iff y= x$ which in sets is just $B_\varepsilon(x)=\{x\}$.
If $\varepsilon > 1$, we have for all $y \in X$ that $d(x,y) \le 1 < \varepsilon$ so $y \in B_\varepsilon(x)$ holds for any $y$ or in sets again $B_\varepsilon(x)=X$.