I would like to prove that an open ball is an open set. I have seen the proof here, but I would like to know exactly why is my proof incorrect.
The definition of an open set in my textbook is the following:
A subset $U$ of a metric space $(X,d)$ is open (in the vector space $X$) if for every $x\in U$, there exists $\epsilon>0$ such that $B(x,\epsilon)\subset U$.
Here's my proof:
Let $B(a,r)=\{x\in X:d(x,a)<r\}$ be an open ball in $X$.
Then for every $x\in B$ there exists $\epsilon \in (0,r]$ such that $B(x,\epsilon)\subset B(a,r)$.
For me this seems to satisfy the definition. But it's wrong probably. Why?
Best Answer
A proof is when you provide convincing reasons/ arguments for the proposition. You're asserted that $\varepsilon$ exists, without giving convincing reasons that it does.
Your "proof" is similar to the following (insufficient) "proof" for why there isn't a greatest real number less than $0$:
This may seem convincing at first, but if you look again, you will see that we have merely asserted that $\ \exists\ y\ $ such that $\ x<y<0.\ $ We haven't actually shown that there exists such a number. So we haven't proven anything. The reason the above "proof" may seem convincing is because we know it works if we fill in the missing part. However, we shouldn't leave out this missing part in our proof. The missing part in the above proof is, for example, $\ y= \frac{1}{3}x.\ $ Then we can write: $\ y-x = \frac{1}{3}x-x = -\frac{2}{3}x > 0\ $ (using the axiom that $\ a.b>0\ $ if $\ a>0\ $ and $\ b>0\ $), therefore $\ y>x.\ $ And $\ y=\frac{2}{3}x<0\ $ because $\ \frac{2}{3}>0\ $ and $\ x<0,\ $ so here we are using the axiom that $\ a.b<0\ $ if $\ a>0\ $ and $\ b<0.\ $ Okay, so now we have actually shown that $\ x<y<0\ $ is always true if we set $\ y=\frac{1}{3},\ $ we can add this into our proof and our proof is fine.
You need to do something similar in your proof that convinces us that you have actually found $\ \varepsilon\ $ such that $\ B(x,\varepsilon)\subset B(a,r).$