Simple Topology Proof

**Aryth** · August 8th 2012, 11:34 AM

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.

**Plato** · August 8th 2012, 11:49 AM

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?

**Aryth** · August 8th 2012, 12:10 PM

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 $U \cup V$ .

**Plato** · August 8th 2012, 12:17 PM

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!

**Aryth** · August 8th 2012, 12:38 PM

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.

**Plato** · August 8th 2012, 12:45 PM

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..

**Aryth** · August 8th 2012, 01:01 PM

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.

**Plato** · August 8th 2012, 01:05 PM

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

**Aryth** · August 8th 2012, 01:07 PM

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

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