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I have a question to the following proof:

Theorem: For $(X,d)$ a metric space, $a \in X$, $r\geq 0$, then the open ball $U_r(a)$ is a open set.

Proof: For $r=0$ is $U_0(a) =\emptyset$ and per defintion is the empty set open. Let $r> 0$, $x \in U_r(a)$ and $\delta:= r-d(x,a) >0$.Than we need to proof that $U_\delta \subseteq U_r(a)$ follows.

Let $y \in U_\delta(x)$ than the Triangle inequality says \begin{align*} d(y,a) \leq d(y,x) + d(x,a) < \delta + d(x,a) = r -d(x,a) + d(x,a) \end{align*} this means $d(y,a) < r$ and therefore $U_\delta \subseteq U_r(a)$. Then $U_r(a)$ is open. $\square$

My question is why can we choose $\delta$? Is it because we know that between two real numbers always is another real number. Or what is the theorem that stands behind that?

Leviathan
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    This looks like a circular argument. You prove that an open ball is open by showing that every point is contained in an open ball within the open ball. By my understanding the topology of a metric space is DEFINED by taking "open" balls as a basis - so they are open by definition. – Tom Collinge Jun 29 '17 at 09:53
  • why $U_\delta(x)\subset U_r(a)$ proves that $U_r(a)$ is open? This is only true if you knows that $U_\delta(x)$ is open... but this is what you are trying to prove. – Masacroso Jun 29 '17 at 11:08
  • @Masacroso that's what I said – Tom Collinge Jun 29 '17 at 11:11
  • But its a common proof, see also here https://math.stackexchange.com/questions/104083/an-open-ball-is-an-open-set – Leviathan Jun 29 '17 at 11:44

3 Answers3

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Since your choice of $\delta$ is $\delta=r-d(x,a)$, then I'd say that we can choose $\delta$ this way because, given two real numbers $x$ and $y$ such that $x>y$, then there is a real number $z>0$ such that $x-y=z$.

José Carlos Santos
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If your question is "why can we choose a $\delta$?"

The definition of an open set $A$ is as follows: for every $a \in A$, there exists a $\delta > 0$ such that $U_\delta(a) \subset A$.

The definition says a $\delta$ exists, so you only have to take 1, so you can choose it.

If your question is "why can we choose that specific $\delta$?"

By definition $d(x, a) < r$, so $r - d(x, a) > 0$ and that is the only condition on $\delta$.

user388557
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Let $ A $ be an open ball in the metric space $ (X,d) $. The, from definition of open ball, we must have $$ A=B_{\varepsilon} (x)=\{ y \in X: d(y,x)<\varepsilon \}$$

Let $ w \in A \iff d(w,x)<\varepsilon \Rightarrow 0<\varepsilon -d(w,x) $

Hence, we choose $ \delta=-d(w,x)+\varepsilon >0 $ and we construct a ball centered at each member of $ A $ which is $ w $ as follows $$B_{\delta} (w)=\{ y \in X: d(y,w)<\delta \}$$ Let $z \in B_{\delta} (w) $. Then $ d(z,w)<\delta=-d(w,x)+\varepsilon \Rightarrow d(z,w)< -d(w,x)+\varepsilon \Rightarrow d(z,w)+d(w,x)<\varepsilon $

From the third axiom of definition of metric space we must have $$d(z,x)\le d(z,w)+d(w,x)<\varepsilon\Rightarrow d(z,x)<\varepsilon \Rightarrow z\in A \Rightarrow B_{\delta} (w) \subset B_{\varepsilon} (w)=A$$

Hence, $ A $ is an open set.

Christian Prince
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