Is the following subset of $\mathbb{R}^2$ closed?

Question

Is the following subset of $\mathbb{R}^2$ closed?

$$A=\left\{(x,\ y)\in\mathbb{R}^2 \mid \left(\frac x3\right)^2+\left(\frac y5\right)^2=1\right\}$$

My solution:

This is an ellipse. We need to check that it contains all its limit points.

For that, we first need to locate the limit points. Consider the euclidean metric $d$ on $\mathbb{R}^2$.

Let $(a,\ b)$ be a point that does not lie on the ellipse. Let the minimum distance from $(a,\ b)$ to $A$ be $\epsilon>0$. Consider $B_d((a,\ b),\ \epsilon)$. This is a neigborhood of $(a,\ b)$ that does not intersect $A$. So, $(a,\ b)$ is not a limit point of $A$.

Let $(c,\ d)$ be a point that lies on the ellipse. I think every neighborhood of $(c,\ d)$ intersects $A$ in some point other than $x$ itself, but I don't know how to prove it.

Please help.

Answer 1

Shorter method - $F(x,y):=(x/3)^2+(y/5)^2$ is smooth(in particular continuous) and $A = F^{-1}(\{1\}).$

Alternatively, roughly in the spirit of your attempt: Suppose $(a_n,b_n)$ is a sequence in $A$ converging in $\mathbb R^2$ to some point $(a,b)$. Then $\lim_{n\to\infty} a_n, \lim_{n\to\infty} b_n$ exist and are equal to $a,b$ respectively, so we are allowed to take $n\to\infty$ in $$ (a_n/3)^2+(b_n/5)^2 = 1$$ to see that $$ (a/3)^2+(b/5)^2 = 1$$ which precisely means that $(a,b) \in A$, as needed.

Alternatively, exactly what you were trying to do: Suppose $\mathbf x$ is a point not on the ellipse. If $d(\mathbf x,A)=0$, then the definition of an infimum allows us to construct a sequence of points in $A$ converging to $\mathbf x$. By the computation in the second proof above, this would imply that $\mathbf x \in A$, a contradiction. Hence $d(\mathbf x,A) > 0$, and $\mathbf x$ is not a limit point of $A$.

If $\mathbf x$ is a point on the ellipse, you can check that $F\in C^1$ satisfies the conditions of the Implicit Function Theorem (or just use the correct formula, depending on $\mathbf x$, out of $a = ± 3\sqrt{1-(b/5)^2}$, $b = ± 5\sqrt{1-(a/3)^2}$) to see that we always have a continuous curve of points in $A$ in any neighbourhood around $\mathbf x$, so that $\mathbf x$ is indeed a limit point of $A$. Hence $A$ is the set of its own limit points, so that $A = \bar A$, giving the result.

Answer 2

A function is continuous is equivalent to inverse images open sets being open. This latter characterization is equivalent to the inverse image of closed sets being closed. Now, the set $\{1\}$ is closed, so its inverse image is closed.

Answer 3

Let the minimum distance from $(a, b)$ to $A$ be $\epsilon>0$. Consider $B_d((a, b), \epsilon)$.

If $d=0$, then $B_d((a,b),\epsilon)$ isn't a neighbourhood of $(a,b)$. In your case $d>0$ is true, but to use this, you already need to know that $A$ is closed. Since you have to prove that, you can't use it.

If you like to argue with limit points, then you have to consider an abritrary sequence $\{(x_n,y_n)\}_{n=1}^\infty$ in $A$ such that $(x_n,y_n)\to (x,y)\in\mathbb R^2$. If you can prove that $(x,y)\in A$, then you are done.

Since $(x_n,y_n)\in A$, we know $$ \left(\frac{x_n}3\right)^2+\left(\frac{y_n}5\right)^2=1 $$ for all $n\in\mathbb N$. Hence the limit becomes $$ \lim_{n\to\infty}\left(\frac{x_n}3\right)^2+\left(\frac{y_n}5\right)^2=1. $$ On the other hand, we can use the rules for convergent sequences and get $$ \lim_{n\to\infty}\left(\frac{x_n}3\right)^2+\left(\frac{y_n}5\right)^2 =\left(\frac{x}3\right)^2+\left(\frac{y}5\right)^2. $$ Combining these two limits yields $$ \left(\frac{x}3\right)^2+\left(\frac{y}5\right)^2=1. $$ Therefore, $(x,y)\in A$.

But in fact, using the method from Calvin Khor is shorter and sufficient if you can use the theorem which is needed.

Answer 4

As many other answers imply, level curves of continuous functions $\mathbb R^2\to\mathbb R$ are closed, being the inverse images of single points (which are closed). Your ellipse is such a set.

Answer 5

Equivalently show that $A^c$ is open.

Let $(x_0,y_0)$ with $f(x_0,y_0) \not =0.$

Assume $f(x_0,y_0) >0.$

f is continuous.

Let $\epsilon >0$.

There is a $\delta >0$ such that

$\sqrt{(x-x_0)^2+(y-y_0)^2} < \delta$ implies

$|f(x,y)-f(x_0,y_0)| \lt \epsilon$, or

$f(x_0,y_0) -\epsilon \lt f(x,y) \lt$

$ f(x_0,y_0) +\epsilon$.

Choose $\epsilon =(1/2)f(x_0,y_0)$ then

$0<(1/2)f(x_0,y_0) < f(x,y)$.

With the corresponding $\delta:$

$(x,y) \in B_{\delta}(x_0,y_0)$ implies

$f(x,y) \gt 0$, i.e. $B_{\delta}(x_0,y_0) \subset A^c$.

Hence $A^c$ is open.

Similarly for $f(x_0,y_0) < 0$.