# Let $M$ be a metric space with the discrete metric or more generally a homeomorph of $M$.

Let $M$ be a metric space with the discrete metric or more generally a homeomorph of $M$.

I've proven that every subset of $M$ is clopen and every function defined on $M$ is continuous.

Let $(x_m)_{m\in\mathbb{R}}$ be a sequence of points defined in metric space $M$, converges to $x\in M$, denoted by that $\lim_{m\to\infty}x_m=x$. Then, the sequences that converge in $M$ would be the ones that take the same value for all large value $m$.

However he said that is wrong, I think this is definitely correct.

I'm currently grading for a class like this. You have to realize that it is completely possible to have exactly the right answer in your head, yet write a proof that is either wrong or so incomplete that we have to give it no points.

You are referencing a good proof. That is, I can take your idea, and make a good proof from it, like so:

Let $X$ be any metric space, and $f: M\to X$ any function. Continuity is equivalent to sequential continuity, so it suffices to show that $f$ is sequentially continuous.

Let $(a_n)$ be a convergent sequence in $M$ with limit $a$. Then, for some $N$, we have $|a_n -a| <1$ for all $n\leq N$. But $M$ has the discrete metric, so $|a_n -a| <1 \implies a_n = a$, and it follows that $(a_n)$ is eventually constant.

Finally, we have $\lim_{n\to \infty} f(a_n) = \lim_{n\to \infty} f(a) = f(a)$. So the sequence $(f(a_n))$ is also convergent, and we conclude that $f$ is sequentially continuous.

But is the proof you actually wrote, actually correct as written? Let's compare:

Let $(x_m)_{m\in\mathbb{R}}$ be a sequence of points defined in metric space $M$, converges to $x\in M$, denoted by that $\lim_{m\to\infty}x_m=x$. Then, the sequences that converge in $M$ would be the ones that take the same value for all large value $m$.

Okay, your proof is shorter (though it's not that much shorter). But it fails to really explain what it's doing, it never mentions continuity or functions at all, it names one sequence then suddenly switches to talking about "the sequences", it uses the vague word "large", and it incorrectly uses $\mathbb{R}$ as an indexing set instead of $\mathbb{N}$ for further confusion.

Finally, it ends with a conclusion about sequences. Sequences! There were no sequences in the problem, so ending your proof with a statement about sequences is confusing even to somebody who understands what you're saying, let alone a person trying to read it for the first time, in the middle of a stack of eighty other problem sets, late at night when his kids want him to read them a story.

The point is that your proof is most certainly not "definitely correct". There's a good idea, but it's just not executed very well, both in technical terms and in your awareness of the reader.

The art of proof is the communication of mathematical ideas to other humans. You must put yourself in their shoes, and understand that there will always be room for improvement in the quality of your mathematical writing. If an instructor does not understand your proof, that is on you, not them.

• Good answer! I just need more experience in writing proofs I guess, are you referring to the class math 104? – user98727 Oct 10 '13 at 4:30
• @user98727 The graduate version. – Slade Oct 10 '13 at 4:59
• Nope I'm a sophomore undergraduate ... I'm just kinda struggle in this class because I never took any analysis class before, and this class goes right into metric space, I do have Pugh, Rudin and Ross but I really don't know how to "utilize" (read) them so I can catch up to the class – user98727 Oct 10 '13 at 5:15
• @user98727 I wasn't saying that you were in a graduate class, just that we're doing similar kinds of material and proofs. And the mistakes are the same at any level. :) Math classes can be very unforgiving if you fall behind... it's a little like falling behind in weightlifting. But I think that working problems thoroughly is almost always the best use of your time, and that reading more than half of the time you spend doing problems is a waste. The only thing better than working problems is hanging out with people who are a little better than you, and not getting intimidated. Good luck! – Slade Oct 10 '13 at 6:20
• Thanks for the tip! I'll keep that in mind! – user98727 Oct 10 '13 at 6:49

The statement is phrased a bit imprecisely, though I’m reasonably sure that you were thinking the right thing. The problem is that for all large values of m is a bit ambiguous, and it isn’t entirely clear that you realize that the necessary degree of ‘largeness’ depends on the individual sequence.

A correct statement is that $\langle x_n:n\in\Bbb N\rangle$ converges if and only if there are an $x\in M$ and an $m\in\Bbb N$ such that $x_n=x$ for all $n\ge m$. Here it’s explicit that the large means $\ge m$ and that $m$ depends on the sequence.

Were you also asked to prove the statement?

• nope the question didn't ask for me to prove it, thanks! – user98727 Oct 10 '13 at 3:51
• @user98727: You’re welcome! – Brian M. Scott Oct 10 '13 at 3:51