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45.35 MB [clear] | game collector pro v5.1.2 te 2013/10/29 12:53…Q:

Showing a sequence is convergent if and only if it converges in every order

Let $x,y,z \in \mathbb{R}$. Show that if $(x_n)_n, (y_n)_n, (z_n)_n \in \mathbb{R}$ is a convergent sequence such that $x_n \geq y_n \geq z_n$ for all $n \geq 0$, then $(x_n)_n, (y_n)_n, (z_n)_n$ is convergent in $\mathbb{R}$.

For part 1
I know how to show that $(x_n)_n, (y_n)_n, (z_n)_n$ converges in $\mathbb{R}$ if $(x_n)_n, (y_n)_n, (z_n)_n$ converges. I just wanted to make sure since I have no clue where to begin for part 2. I really don’t know how to approach this problem. Can anyone help me? Thanks in advance.


Your claim follows directly from

Theorem. If $(a_n)_n, (b_n)_n, (c_n)_n$ is a convergent sequence, then $(a_n,b_n,c_n)_n$ is convergent iff each convergent subsequence of $(a_n,b_n,c_n)_n$ has convergent subsequence.

Note that it does not matter whether we consider $(a_n)_n, (b_n)_n, (c_n)_n$ to be convergent in $\mathbb{R}$, or in $\mathbb{R}^2$, or in $\mathbb{R}^3$.
Now let $(x_n)_n, (y_n)_n, (z_n)_n$ be a convergent sequence in $\mathbb{R}$, and suppose that $(x_n)_

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