Suppose that $x_n \rightarrow x$ as $ n \rightarrow \infty$. Prove from first principles that if $x_n \leq M$ for all $n$, then $x \leq M$.

Question

I've been asked to try my hand at the above proof. I'm relatively new to propositional logic and analysis so any tips you could offer would be much appreciated!

I'm going to try prove by contradiction.

Take the statement.

$$ \forall n((x_n \leq M)\implies(x \leq M))$$

Then negate.

$$ \neg(\forall n((x_n \leq M)\implies(x \leq M)))$$

$$ \exists n((x_n \leq M) \land (x > M))$$

Here we find our contradiction there is no value of $n$ in which the sequence $x_n$ and its limit $x$ can both be equal to or less than $M$ and strictly greater than $M$. There for contradiction and the above statement is true.

Answer 1

Here is a hint. On the contrary, suppose $x > M$. Then $x - M > 0$. Thus, in the definition of sequential convergence, we can take $\epsilon = x - M$. Now write down the rest of the definition of $x_n \to x$ and try to derive a contradiction. If you'd like me to elaborate more, let me know.

Answer 2

Use the definition of $x_n \to x$.

For any $\epsilon > 0$ then there is an $N$ so that $n>N\implies |x_n - x| < \epsilon$.

Now if we assume $x > M$ then we have $|x_n - x| < \epsilon$ so $-\epsilon < x-x_n < \epsilon$ so $x_n - \epsilon < x < x_n + \epsilon$. And we also have $M < x$ and $x_n < M$ so we have $x_n - \epsilon < x_n < M < x < x_n + \epsilon$.

Can we get a contradiction from that? Hint: We can always for such an $x_n$ for any $\epsilon$. Just how much bigger can $x$ be than $M$?