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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.

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  • $\begingroup$ That's not a valid proof. All you have done is negated the statement you're trying to prove. You should try something along the lines of $|x| = |x -x_n + x_n| = ...$ Can you take it from there? $\endgroup$ Apr 18, 2020 at 19:14
  • $\begingroup$ You must show that the negation of the statement you are trying to prove, actually leads to contradiction. As Mr. Nicholas has mentioned, you have only negated the statement which is to be proven. $\endgroup$
    – Koro
    Apr 18, 2020 at 19:20
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    $\begingroup$ "Here we find our contradiction there is no value of n in which the sequence xn and its limit x can both be equal to or less than M and strictly greater than M" Why not? Why can't we have all $x_n < 27$ but have $x_n \to 28$? Why not? Just saying you can't doesn't prove anything. You have to show you can't. YOu prove comes down to seeing "Assume it is false. Well then it is clearly false. So we have a contradiction." $\endgroup$
    – fleablood
    Apr 18, 2020 at 19:52
  • $\begingroup$ After all we can have all $x_n = 1-\frac 1n$ and then $x_n < 1$ but if $x_n\to x=1$. But $x = 1$ and $x \not < 1$. Is that a contradiction? No, it is not. So why is $x_n < M$ and $x_n \to x$ and $x > M$ and $x \not < M$ a contradiction? $\endgroup$
    – fleablood
    Apr 18, 2020 at 19:58

2 Answers 2

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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.

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  • $\begingroup$ Thanks for the hint! I've tried a solution using your hint. Based on above we can say $|x_n - x| < x -M$, squaring both sides we get, $x_n^2 - 2x_n x < M^2 - 2xM$. $x_n^2 \leq M^2$ and $-2x_n x \leq -2xM$. Here is a contradiction as a case, based on $x_n \leq M < x$ exists a case where $|x_n - x| = x-M$ . If I am still incorrect (which is likley) please elaborate I'd love to see the proper way! $\endgroup$
    – UmamiBoy
    Apr 19, 2020 at 20:03
  • $\begingroup$ It's unclear what you're contradicting. I think a simpler, cleaner approach would be to rewrite the expression $|x_n - x| < x - M$ without absolute value bars. From this, you can obtain that $x_n > M$ for all $n > N$. This is a contradiction since we know that $x_n \leq M$ for all $n$. $\endgroup$ Apr 19, 2020 at 20:29
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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 $M$?

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