# Math Help - Simple Topology Proof

1. ## Simple Topology Proof

Suppose $U$ and $V$ are pointsets of $\mathbb{R}$. If $U$ and $V$ both have a leftmost point, prove that the union $U \cup V$ also has a leftmost point.

I can kind of see how this would be true, but I am having a lot of trouble formalizing a proof. If someone could help me along that would be great. I have just started this course, and it's a little difficult to get used to.

2. ## Re: Simple Topology Proof

Originally Posted by Aryth
Suppose $U$ and $V$ are pointsets of $\mathbb{R}$. If $U$ and $V$ both have a leftmost point, prove that the union $U \cup V$ also has a leftmost point.
Suppose that $\alpha$ is the left most point of $U$. That means that $\alpha\in U$ and $\forall x\in U$ we have $\alpha\le x~.$.
Make similar about $\beta$ the left most point of $V$.

Clearly both $\alpha~\&~\beta$ belong to $U\cup V$.

Can you finish?

3. ## Re: Simple Topology Proof

Hmmm... So... Just do a similar construction?

Since $\alpha \ \& \ \beta$ belong to $U \cup V$, then $\forall x \in U \cup V$ either $\alpha \leq x$ or $\beta \leq x$, which guarantees that there is a leftmost point in $U \cup V$.

4. ## Re: Simple Topology Proof

Originally Posted by Aryth
Hmmm... So... Just do a similar construction?

Since $\alpha \ \& \ \beta$ belong to $U \cup V$, then $\forall x \in U \cup V$ either $\alpha \leq x$ or $\beta \leq x$, which guarantees that there is a leftmost point in $U \cup V$.
No you are not done.
There is only one left most point of $U\cup V.$

Now it is one of $\alpha\text{ or }\beta.$ Now we know that $$\left( {\alpha = \beta {\kern 1pt} } \right){\kern 1pt} \vee \left( {{\kern 1pt} \alpha < \beta } \right){\kern 1pt} {\kern 1pt} \vee \left( {{\kern 1pt} \alpha < \beta } \right){\kern 1pt} {\kern 1pt}$$.

Finish it!

5. ## Re: Simple Topology Proof

I'm slightly confused... I know what you're saying, but I can't see the rest of the proof... I mean, I could suppose that $\alpha \leq \beta$ and that would make $\alpha$ the left most point, or I could say the opposite, and $\beta$ would be the left most point. I feel like I'm missing something.

6. ## Re: Simple Topology Proof

Originally Posted by Aryth
I'm slightly confused... I know what you're saying, but I can't see the rest of the proof... I mean, I could suppose that $\alpha \leq \beta$ and that would make $\alpha$ the left most point, or I could say the opposite, and $\beta$ would be the left most point. I feel like I'm missing something.
You know $$\left( {\alpha = \beta {\kern 1pt} } \right){\kern 1pt} \vee \left( {{\kern 1pt} \alpha < \beta } \right){\kern 1pt} {\kern 1pt} \vee \left( {{\kern 1pt} \alpha < \beta } \right){\kern 1pt} {\kern 1pt}$$.

Take each case. Prove that $\alpha\text{ or }\beta$ is the left most point..

7. ## Re: Simple Topology Proof

Ok...

Note: If $x \in U \cup V$, then $\left[x \in U \vee x \in V \vee (x \in U \wedge x \in V)\right]$. In any of these cases, either $\alpha$ or $\beta$ is the left most point.

We know: $$\left( {\alpha = \beta {\kern 1pt} } \right){\kern 1pt} \vee \left( {{\kern 1pt} \alpha < \beta } \right){\kern 1pt} {\kern 1pt} \vee \left( {{\kern 1pt} \beta < \alpha } \right){\kern 1pt} {\kern 1pt}$$

Let $\alpha = \beta$, then they are the same point, and that point is the left most point, since it is also the left most point of both $U$ and $V$.

Let $\alpha < \beta$, then $\forall x \in U$ and $\forall x \in V$, $\alpha \leq x$. In other words, it means that $\alpha$ is to the left of all points in $V$, and is therefore the left most point of the union.

Let $\beta < \alpha$, then $\forall x \in U$ and $\forall x \in V$, $\beta \leq x$. In other words, it means $\beta$ is to the left of all points in $U$, and is therefore the left most point of the union.

8. ## Re: Simple Topology Proof

Well done. It is now complete.
I think it is generally true that topologist are sticklers for completeness in proofs .

9. ## Re: Simple Topology Proof

I really appreciate the help. And I'll keep that in mind, I'm getting that feeling from my professor.