Proof: open ball is an open set

[nuuskur]

Homework Statement

Second try at the proof, this time with correct vocabulary, I hope
Prove that an open ball is an open set.

Homework Equations

The Attempt at a Solution

Let B(P_0, r) be an open ball in \mathbb{R}^m, where P_0 is the centerpoint of the ball and r > 0 is its radius.
Assume point P\in B(P_0,r). Our objective is to show that every point P in the open ball is an interior point \Leftrightarrow The open ball is an open set.
Therefore \exists\varepsilon > 0\colon B(P,\varepsilon)\subset B(P_0, r). Assume also point S\in B(P,\varepsilon).
Per the triangle inequality we know: \forall S\in B(P,\varepsilon) \Rightarrow d(S,P_0)\leq d(S,P) + d(P,P_0)<\varepsilon + d(P,P_0).

Fix \varepsilon\colon = r - d(P,P_0) > 0 then \forall S\in B(P,\varepsilon);
d(S,P_0) < \varepsilon + d(P,P_0) = r, therefore every point in the open ball is an interior point and the open ball is an open set._{\blacksquare}

 

[HallsofIvy]

Homework Statement

Second try at the proof, this time with correct vocabulary, I hope
Prove that an open ball is an open set.

Homework Equations

The Attempt at a Solution

Let B(P_0, r) be an open ball in \mathbb{R}^m, where P_0 is the centerpoint of the ball and r > 0 is its radius.
Assume point P\in B(P_0,r). Our objective is to show that every point P in the open ball is an interior point \Leftrightarrow The open ball is an open set.
Therefore \exists\varepsilon > 0\colon B(P,\varepsilon)\subset B(P_0, r).

No, This is what you want to prove.

Assume also point S\in B(P,\varepsilon).
Per the triangle inequality we know: \forall S\in B(P,\varepsilon) \Rightarrow d(S,P_0)\leq d(S,P) + d(P,P_0)<\varepsilon + d(P,P_0).

Fix \varepsilon\colon = r - d(P,P_0) > 0 then \forall S\in B(P,\varepsilon);
d(S,P_0) < \varepsilon + d(P,P_0) = r, therefore every point in the open ball is an interior point and the open ball is an open set._{\blacksquare}

Your proof doesn't work because you started by assuming what you want to prove. Instead, note that since P is in B(P_0,r), d(P,P_0)< r so 0< r- d(P,P_0). Use that to determine a specific value for \epsilon[/tex]

 

[nuuskur]

I am confused. Where have I made a mistake?
If the distance between S and P₀ is always strictly less than r itself, then every S is an interior point in the ball and as S itself is a point in the ball around point P, then that automatically renders every P to be an interior point. Is that not correct?

EDIT: oh I think I understand. Game of words.

Let B(P_0, r) be an open ball in \mathbb{R}^m, where P_0 is the centerpoint of the ball and r > 0 is its radius.
Assume point P\in B(P_0,r). Our objective is to show that every point P in the open ball is an interior point \Leftrightarrow The open ball is an open set.
Therefore \exists\varepsilon > 0\colon B(P,\varepsilon)\subset B(P_0, r). Note that \forall\varepsilon > 0,\forall S\in B(P,\varepsilon) \Rightarrow d(S,P_0)\leq d(S,P) + d(P,P_0)<\varepsilon + d(P,P_0).

Fix \varepsilon\colon = r - d(P,P_0) > 0 then \forall S\in B(P,\varepsilon);
d(S,P_0) < \varepsilon + d(P,P_0) = r, therefore every point in the open ball is an interior point and the open ball is an open set._{\blacksquare}

The triangle inequality holds indefnitely, I'll fix a certain epsilon around that fact and show that each and every S is an interior point in the big ball which also means any P will be an interior point because of that.

 

[micromass]

I am confused. Where have I made a mistake?
If the distance between S and P₀ is always strictly less than r itself, then every S is an interior point in the ball and as S itself is a point in the ball around point P, then that automatically renders every P to be an interior point. Is that not correct?

EDIT: oh I think I understand. Game of words.

Let B(P_0, r) be an open ball in \mathbb{R}^m, where P_0 is the centerpoint of the ball and r > 0 is its radius.
Assume point P\in B(P_0,r). Our objective is to show that every point P in the open ball is an interior point \Leftrightarrow The open ball is an open set.

Not a serious mistake, but you can't say "take a point $P$" and then "we need to prove all points $P$ are interior points". The thing is that if you say "take a point $P$", then I assume that you have taken a fixed point in the ball, and that point does not change to some other points in the rest of the proof. But then you say "we need to prove all points $P$ are interior points", which seems to say that your point $P$ can still be any point.

It's not a big mistake obviously, but it's important to write proofs in a logical way. So what you should say "Assume a point $P\in B(P_0,r)$. Our objective is to show that $P$ is an interior point.

Therefore \exists\varepsilon > 0\colon B(P,\varepsilon)\subset B(P_0, r).

You say "therefore", so it seems like this follows from the previous. But I don't see how this follows from the previous statements. This is what you need to prove.

Note that \forall\varepsilon > 0,\forall S\in B(P,\varepsilon) \Rightarrow d(S,P_0)\leq d(S,P) + d(P,P_0)<\varepsilon + d(P,P_0).

Fix \varepsilon\colon = r - d(P,P_0) > 0 then \forall S\in B(P,\varepsilon); d(S,P_0) < \varepsilon + d(P,P_0) = r

OK, but you can't do this. I understand this is how you found the proof, and that's ok. But if you present the proof then you can't just work with a $\varepsilon$ and then fix it later (only in some extreme rare situations does this happen). So you need to say first that you define $\varepsilon = r-d(P,P_0)$. Then you need to prove that $\varepsilon>0$and that with this choice of $\varepsilon$ we have $B(P,\varepsilon)\subseteq B(P_0,r)$.

Other remarks:
Please do not use the following symbols in your proofs ever: $\forall$, $\exists$, $\Rightarrow$, $:$ This might seem strange to you. After all, they have taught you to do proofs this way. But in fact, it is an unspoken agreement under many mathematicians that those symbols make the proof less readable. Just type out the words "for every", etc. Every sentene in a proof must be a sentence that is easily readable. See http://www.math.washington.edu/~lee/Writing/writing-proofs.pdf

So your proof wasn't bad, but the problems were more in the presentation.