Re: Simple Topology Proof

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

Suppose that $\displaystyle \alpha$ is the *left most* point of $\displaystyle U$. That means that $\displaystyle \alpha\in U$ and $\displaystyle \forall x\in U$ we have $\displaystyle \alpha\le x~.$.

Make similar about $\displaystyle \beta$ the *left most* point of $\displaystyle V$.

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

Can you finish?

Re: Simple Topology Proof

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

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

Re: Simple Topology Proof

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

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

**No you are not done.**

There is only __one__ *left most* point of $\displaystyle U\cup V.$

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

Finish it!

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 $\displaystyle \alpha \leq \beta$ and that would make $\displaystyle \alpha$ the left most point, or I could say the opposite, and $\displaystyle \beta$ would be the left most point. I feel like I'm missing something.

Re: Simple Topology Proof

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**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 $\displaystyle \alpha \leq \beta$ and that would make $\displaystyle \alpha$ the left most point, or I could say the opposite, and $\displaystyle \beta$ would be the left most point. I feel like I'm missing something.

You know $\displaystyle \[\left( {\alpha = \beta {\kern 1pt} } \right){\kern 1pt} \vee \left( {{\kern 1pt} \alpha < \beta } \right){\kern 1pt} {\kern 1pt} \vee \left( {{\kern 1pt} \alpha < \beta } \right){\kern 1pt} {\kern 1pt} \]$.

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

Re: Simple Topology Proof

Ok...

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

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

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

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

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

Re: Simple Topology Proof

Well done. It is now complete.

I think it is generally true that topologist are sticklers for completeness in proofs .

Re: Simple Topology Proof

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