Assume $X,Y \subset \Bbb R$ and both have the Bolzano-Weierstrass Property. Prove that $X \cup Y$ has the Bolzano-Weierstrass Property.

Question

I'm not quite sure if my proof is complete:

PF -

• Let $\{x_n\}$ be a sequence of X where $\lim_{n \rightarrow \infty} x_n = L_1 \in E$
• Let $\{y_n\}$ be a sequence of Y where $\lim_{n \rightarrow \infty} y_n = L_2 \in E$
• Thus we know $\{x_n\}$ is a sequence of $X \cup Y$ since $X \subset X \cup Y$ and $\{y_n\}$ is a sequence of $X \cup Y$ since $Y \subset X \cup Y$
• Sequences in $X \cup Y$ are convergent because X and Y have the BWP property
• We know $x_n \rightarrow L_1\in E$ and $y_n \rightarrow L_2\in E$ and that $X \cup Y$ must contain $\{x_n\}$ or $\{y_n\}$, so we know that the sequences in $X \cup Y$ converge to points in E
• Therefor $X \cup Y$ has the BWP property

Answer 1

You are not proving what you are asked to show. You should start with an arbitrary sequence $\{z_n\}$ that is contained in $X\cup Y$, and then prove that it has a subsequence that converges to a point in $X\cup Y$. What you showed is that if you take a sequence in $X\cup Y$ that is wholly contained in $X$, or wholly contained in $Y$, then it has a subsequence converging to a point in $X$, respectively $Y$.

So, start with a sequence $\{z_n\}$, with $z_n\in X\cup Y$ for each $n$. There are two cases to consider:

1. There exists $N$ such that for all $n> N$, $z_n \in X$, or
2. There is a subsequence $\{n_k\}$ of $\{n\}$ such that $z_{n_k} \in Y$ for each $k=1,2,3,\dots$.

In case 1, apply the Bolzano-Weierstrass property of $X$ to the sequence $\{x_k\}$ defined by $x_k := z_{N+k}$, $k = 1,2,3,\dots$.

In case 2, apply the Bolzano-Weierstrass property of $Y$ to the sequence $\{y_k\}$ defined by $y_k := z_{n_k}$, $k = 1,2,3,\dots$.

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About the two cases.

The first case formalizes what happens if the sequence $z_n$ eventually lies in $X$ beyond some time step $N$. The second case says that if the sequence does not eventually lie in $X$, then it must hop into $Y$ for infinitely many time steps, since each point $z_n$ belongs to either $X$ or $Y$. The cases have some overlap if $X$ and $Y$ are not disjoint, but the thing that matters is that we cover all the cases, which we can check:

Suppose that 1 does not hold. Then for each $N$, there is some $n_N > N$ such that $z_{n_N}\notin X$. Moreover, we can arrange for $n_N < n_{N+1}$ for each $N$. But then $z_{n_N}$, $N=1,2,3,\dots$ is a subsequence contained in $Y$, so 2 holds. You can check similarly that if 2 does not hold, then 1 must hold.