Help with a proof regarding convergent sequence (proof by contradiction)

[malawi_glenn]

Ok I am trying to brush up my real analysis skills so that I can study some topology and measure theory at some point.

I found this theorem in my notes, that is proven by using proof by contradiction. However, I have a hard time understanding what the contradiction really is...

Here is the theorem and the proof.
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Let $(x_n)$ be a convergent sequence. Assume that there exists a natural number $N$ such that $x_n \leq A$ (where $A$ is some real number) holds for each $n \geq N$. Then the following holds: $\lim_{n \to \infty} x_n \leq A$.

Proof: Assume the contrary, that $\lim_{n \to \infty} x_n = L > A$. Let $\epsilon = L - A$. Since $(x_n)$ is convergent, all except a finite number of elements in the sequence belongs on the interval $(L - \epsilon , L + \epsilon)$. But all numbers on this interval is strictly greater than $A = L - \epsilon$, which is a contradiction.
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I have no idea what is contradicted and why this is a contradiction. Thanks in advance for any kind of illumination on this theorem and proof.

 

[Gaussian97]

The contradiction is that your initial statement is $x_n \leq A$ (except maybe a finite number of cases). But if you assume $\lim x_n > A$, then you need an infinite number of cases where $x_n > A$.
This is a contradiction and therefore you must have $\lim x_n \leq A$

 

[PeroK]

Ok I am trying to brush up my real analysis skills so that I can study some topology and measure theory at some point.

I found this theorem in my notes, that is proven by using proof by contradiction. However, I have a hard time understanding what the contradiction really is...

Here is the theorem and the proof.
- - - - - - - - - - - - - - - - - - - - - - - -
Let $(x_n)$ be a convergent sequence. Assume that there exists a natural number $N$ such that $x_n \leq A$ (where $A$ is some real number) holds for each $n \geq N$. Then the following holds: $\lim_{n \to \infty} x_n \leq A$.

Proof: Assume the contrary, that $\lim_{n \to \infty} x_n = L > A$. Let $\epsilon = L - A$. Since $(x_n)$ is convergent, all except a finite number of elements in the sequence belongs on the interval $(L - \epsilon , L + \epsilon)$. But all numbers on this interval is strictly greater than $A = L - \epsilon$, which is a contradiction.
- - - - - - - - - - - - - - - - - - - - - - - -

I have no idea what is contradicted and why this is a contradiction. Thanks in advance for any kind of illumination on this theorem and proof.

It's quite neat to do it that way. Can you do a different proof? Along similar lines.

 

[FactChecker]

I think that you should fill in some details in a more step-by-step manner.
Let $N_1$ denote the integer such that $|x_n -L| \lt \epsilon$ $\forall n\gt N_1$. Let $N' = max\{N_1, N\}$. If $n \gt N'$, what can you say about $x_n$?
Although this amount of detail might seem unnecessary, it is good practice for problems that are more complicated.

 

[PeroK]

There's another useful technique, as follows:

Let $\epsilon > 0$. Show that $L \le A + \epsilon$. Then use the result (you could prove this separately) that: $$\text{If} \ \forall \ \epsilon > 0 \ \text{we have} \ a \le b + \epsilon, \ \text{then} \ a \le b$$