1. ## Proof linking continuity and convergence

i have no background in analysis, and have absolutely no idea how to write proofs, need some help proving this theorem using epsilon:

if sequence x (x_n) approaches "a", and sequence y (y_n) approaches b and sequence x (x_n) is greater than or equal to sequence y (y_n) for all n, then "a" is greater than or equal to "b".

thanks

2. Originally Posted by ml692787
i have no background in analysis, and have absolutely no idea how to write proofs, need some help proving this theorem using epsilon:

if sequence x (x_n) approaches "a", and sequence y (y_n) approaches b and sequence x (x_n) is greater than or equal to sequence y (y_n) for all n, then "a" is greater than or equal to "b".

thanks
If look here.

As a result we have a simple result.

Theorem 1: Let $(x_n)$ be a convergent sequence of non-negative terms. Then its limits is non-negative. (This should be obvious from the link I gave you).

Theorem 2: Let $x_n\geq y_n$ for all $n$. And $(x_n) \mbox{ and }(y_n)$ are convergent sequences. Then $\lim \ x_n \geq \lim \ y_n$.

Proof: Define a sequence $z_n = x_n - y_n$. Now this means that $z_n \geq 0$. And furthermore $(z_n)$ is convergent since it is a difference of two convergent sequences. Thus, $\lim \ z_n \geq 0$. Thus, $\lim \ x_n - \lim \ y_n \geq 0$ because the limit of $z_n$ is the difference of the two individial sequences. Thus, $\lim \ x_n \geq \lim \ y_n$ by Theorem 1. Q.E.D.

3. Originally Posted by ThePerfectHacker
If look here.

As a result we have a simple result.

Theorem 1: Let $(x_n)$ be a convergent sequence of non-negative terms. Then its limits is non-negative. (This should be obvious from the link I gave you).

Theorem 2: Let $x_n\geq y_n$ for all $n$. And $(x_n) \mbox{ and }(y_n)$ are convergent sequences. Then $\lim \ x_n \geq \lim \ y_n$.

Proof: Define a sequence $z_n = x_n - y_n$. Now this means that $z_n \geq 0$. And furthermore $(z_n)$ is convergent since it is a difference of two convergent sequences. Thus, $\lim \ z_n \geq 0$. Thus, $\lim \ x_n - \lim \ y_n \geq 0$ because the limit of $z_n$ is the difference of the two individial sequences. Thus, $\lim \ x_n \geq \lim \ y_n$ by Theorem 1. Q.E.D.
Hey, TPH, could we have said the following?

Since the sequence $\{ x_n \}$ approaches $a$, we have $\lim_{n \to \infty} x_n = a$. Similarly, since $\{ y_n \}$ approaches $b$, we have $\lim_{n \to \infty} y_n = b$

Now, we have $x_n \geq y_n$ for all $n$

$\Leftrightarrow \lim_{n \to \infty}x_n \geq \lim_{n \to \infty}y_n$

$\Leftrightarrow a \geq b$

as desired

4. Originally Posted by Jhevon
Hey, TPH, could we have said the following?

Since the sequence $\{ x_n \}$ approaches $a$, we have $\lim_{n \to \infty} x_n = a$. Similarly, since $\{ y_n \}$ approaches $b$, we have $\lim_{n \to \infty} y_n = b$

Now, we have $x_n \geq y_n$ for all $n$

$\Leftrightarrow \lim_{n \to \infty}x_n \geq \lim_{n \to \infty}y_n$

$\Leftrightarrow a \geq b$

as desired
But you are accepting a theorem that we are trying to prove!

For example, this is what you are doing. Say we want to be that the sum of the triangles add up 180 degrees (let us called this Euclid's Principle). We prove Euclid's principle using the Parallel Postulate. Then you want to justify the Parallel Postulate and you use Euclid's principle. How is that a proof?