Open or closed set?

Question

$M := \lbrace (x,y)^T \in \mathbb{R}^2 \mid x^9+y^9 \leq 25 \rbrace $

Is M open or closed?

I guess its closed and open at the same time.

Answer 1

Your guess is half-correct:

• It is closed, because $M=f^{-1}\bigl((-\infty,25]\bigr)$, where $f$ is the continuous function defined by $f(x,y)=x^9+y^9$ and $(-\infty,25]$ is closed.
• It is not open, because $\left(\sqrt[9]{25},0\right)\in M$, but no open ball centered at $\left(\sqrt[9]{25},0\right)$ is contained in $M$.

Answer 2

Being the preimage by a continuous function of a closed set, your set is closed. Being $\mathbb{R}^2$ connected, its only open and closed sets are $\emptyset$ and $\mathbb{R}^2$. Your set isn't $\emptyset$ because $(0,0)$ belongs to it, while it isn't $\mathbb{R}^2$ because $(1000,0)$ doesn't belong to it. So, your set, being closed, can't be open.

Answer 3

For fun, consider :

$B:= ${$(x,y)| x^9+y^9>25$}.

$B=f^{-1}((25,\infty))$ as the preimage of the continous function $f(x,y):= x^9+y^9$ of an open set is open.

$M \cap B= \emptyset$, and $M \cup B =\mathbb{R^2}$.

Hence $M= \mathbb{R^2}$ \ $B$ is closed.

Since $M \not = \emptyset$ , and $M \not =\mathbb{R^2}$, $M$ is not open.