I have a question.
I have to check whether $I= \begin{Bmatrix} x \in R^3 : 1 \leq x_2 \leq 3,0 \leq x_2,−1 \geq x_3 \end{Bmatrix}$ is an open set, a closed set or neither. I thought I want to show it with an open disc (open ball), will someone help me?
Thank you
Hint:Take a point which is in the complement of your set, there is some condition that is not met, show that you can find a small ball around that point that violates the same condition and deduce that the entire ball is contained in the complement
$$I = \begin{Bmatrix} x \in R^3 : 1 \leq x_1 \leq 3,0 \leq x_2,−1 \geq x_3 \end{Bmatrix}$$ If we take a disc around any point with $x_2 = 0$, the point is in $I$, but the disc contains points with $x_2 < 0$, which are not in $I$. So $I$ is not an open set.
To check whether $I$ is closed, we should check if its complement is open. We have $$I^C = \begin{Bmatrix} x \in R^3 : x_1 < 1\ \vee x_1 > 3\ \vee x_2 < 0\ \vee −1 < x_3 \end{Bmatrix}$$ A point is in $I^C$ if it satisfies at least one of the four inequalities. If it satisfies $x_1 < 1$, then you can use for that point an open disc of radius $\epsilon = 1 - x_1$. Every point $y$ in that disk has $2x_1-1 < y_1 < 1$, so $y_1 < 1$, so every point in the disk satisfies the first inequality and is also in $I^C$. You can do the same for the other inequalities. So $I^C$ is an open set, so $I$ is a closed set.