7

I am trying to prove that the set $A=\{(x,y)|x^2\leq y\}$ is closed in $\mathbb R^2$. I wrote a proof, but I think the end is wrong. My proof is:

Consider the set $A=\{(x,y)|x^2\leq y\}$ in $\mathbb R^2$. Let $(x_n,y_n)_{n \in \mathbb N}$ be a sequence in $A$ that converges to $(x,y) \in \mathbb R^2$. By the componentwise convergence criterion, $(x_n,y_n) \to (x,y)$ iff $x_n \to x$ and $y_n \to y$ as $n \to \infty$.

(From here on out, I think it is wrong as I really didn't know how to proceed). So, $x_n^2 \to x^2$ and $y_n \to y$ as $n \to \infty$. Since $\forall n \in \mathbb N \ \ (x_n,y_n)\in A$, $x_n^2\leq y$ $\forall n\in \mathbb N$. Hence by taking the limit as $n \to \infty$ of both sides, $x^2\leq y$. Therefore, $(x,y)\in A$ and so $A$ is closed.

CleoFroPollo
  • 111
  • 1
    Your proof is absolutely correct. Only it is $x^2\leq y$ of course, not $x^2\to y$. And just before that I guess you wanted to write $x_n^2\leq y_n\forall n\in\mathbb{N}$. But the mathematical part of the proof is fine. – Mark Feb 17 '19 at 21:09
  • Oh, of course! Sorry, looks like I had a typo. Thanks! – CleoFroPollo Feb 17 '19 at 21:12
  • Yeah, this is what I thought. The proof is very good. – Mark Feb 17 '19 at 21:13
  • you might want to say something about the fact that the function $x \mapsto x^2$ is continuous, that's really the heart of the proof (or at least the sentence "so $x_n^2 \to x^2$ – Andres Mejia Feb 17 '19 at 21:42
  • @Andres Mejia, you don't need continuity here. It is a basic property of convergent sequence: if $a_n\to a,b_n\to b$ then $a_nb_n\to ab$. If anything, this is one of the ways to prove the continuity of $x^2$. (of course if in this exercise you say that $x_n^2\to x$ because you know $x\to x^2$ is continuous then it is also correct) – Mark Feb 17 '19 at 21:47
  • @Mark your "simpler" statement is nothing but the continuity of $(x,y) \mapsto xy$. If anything, proving continuity of $x \mapsto x^2$ is more elementary. – qualcuno Feb 17 '19 at 21:50
  • @Guido A. it depends. When I took my first course in calculus we started from learning about limits of sequences and only in the second half of the course we started talking about limits of functions. So we did the hard work with sequences, and then in that second half we found out that we don't have to prove almost any properties about limits of functions, because they all trivially come from sequences. So I knew the properties of limits of sequences a few week before I even started learning about continuity. – Mark Feb 17 '19 at 21:53
  • @Andres Mejia What I'm saying is that you can prove these properties of limits of sequences without even knowing the definition of continuity. This is exactly how it was in my first calculus course. – Mark Feb 17 '19 at 22:00
  • well, yeah... most of these properties are subsumed by the notion of sequential continuity. Anyway, I now feel that I am just bickering. My point was that the OP should say something in that sentence-- how it is phrased is up to the writer – Andres Mejia Feb 17 '19 at 22:06
  • Another way is to show that the complement is open, that is if $x^2 >y$ then there is an open set containing $(x,y)$ such that $(x')^2 > y$ for all $(x',y')$ in the open set. – copper.hat Feb 18 '19 at 19:37

2 Answers2

4

Your argument seems fine, although a bit hard to read (if you allow me to get nit-picky, of course this is a matter of style). Here's a possible clarification of your idea: since for each $n \in \mathbb{N}$ we have that $(x_n,y_n) \in A$, then $x_n^2 \leq y_n$. Hence

$$ 0 \leq y_n-x_n^2 \quad (\forall n \geq 1). $$

Taking limits when $n \to \infty$ and using convergence of each sequence, we get that $0 \leq y-x^2$ (here we use that $t \mapsto t^2$ is continuous). Therefore we see that $x^2 \leq y$, or equivalently, $(x,y) \in A$.

qualcuno
  • 17,121
  • Okay, thanks. I do have another question, why is it that this argument does not work when you have $A={(x,y)|x^2>y}$? – CleoFroPollo Feb 17 '19 at 21:19
  • 2
    The main issue is that in some intuitive sense, limits do not behave well with respect to strict inequalities. For example, if you have a sequence $(a_n)_n$ such that $0 < a_n$ for all $n$, then you can only conclude that $0 \leq \lim_n a_n$, losing the strict bound that you had term to term (prove this if you haven't!). A counterexample for the 'strict inequality result' can be $(\frac{1}{n})_n$ which clearly is of positive terms, but converges to zero. – qualcuno Feb 17 '19 at 21:23
  • Ohh, right! Thanks so much. – CleoFroPollo Feb 17 '19 at 21:25
0

Let $f: (x,y) \mapsto x^2-y$, now we can clearly see that $A = f^{-1}[\mathbb R_-]$. $f$ is continuous as it is a polynomial.

We've shown that $A$ is closed, being the inverse image of closed set $\mathbb R_-$ by continuous function $f$.

anni
  • 919