I am starting to study topology, and to assess if I am on the right track, I kindly ask if someone can check my reasoning below.

  1. Let the metric space $(\mathbb{R}^2,d)$, where $d$ is the Euclidean distance, and let $X=\{x\in \mathbb{R}^2:0 \le x_1 < 1 \}$. To see that $X$ is not open, it suffices to take a ball centered on any point with $x_1=0$ and observe that the points with coordinate $x_1 <0 $ belongs to the ball but not to $X$. Therefore, $X$ is not open since the whole ball is not entirely contained $X$. Is that correct?

  2. Consider now the metric space $(X,d)$, where $X$ is the same as defined in 1. and $d$ is the Euclidean distance. Let $A=\{x\in \mathbb{R}^2: x_1^2+x_2^2 \le 2\}$. As it may seem strange, the segment $S=\{ x\in \mathbb{R}^2: x_1=0, -1 \le x_2 \le 1\}$ belongs to the interior of $A$ for the following reason: by definition of ball, one shall consider only the points contained in $X$, otherwise the considered set is not a ball, (even if it has a typical "round shape"). In-fact, by taking any point on the segment $S$ as the center of any ball, then all the points with coordinate $x_1<0$ of a hypothetical "rounded shape" ball are ruled out and shall not be considered. Is that correct?

  3. Let $(X,d)$ and $(Y,d)$ two metric spaces with $X=\mathbb{R}^3 $ and $Y=\{x \in \mathbb{R}^3: x_1 \le 1, x_2\le 1, x_3=0\}$. Let $d$ be the Euclidean distance for both the spaces. Finally, let $A=\{x\in \mathbb{R}^3: 0<x_1<1, 0<x_2\le 1, x_3 =0\}$. The interior set of $A$ with respect to $(X,d)$ is the empty set. This is because every ball contained in $A$ has some point with $x_3 \neq 0$ which are not contained in $A$, but are contained in $X$. On the other hand, by considering the metric space $(Y,d)$, then the interior of $A$ is $A^°=\{x\in \mathbb{R}^3 :0<x_1<1, 0<x_2< 1, x_3 =0\ \}$. This because, by definition of ball, it shall be entirely contained in $Y$, and in this case a ball in $Y$ must necessarily have $x_3=0$ (otherwise it is not a ball). The key point in obtaining different results in this exercise is due to the different "ball shape" when considering the metric space $(X,d)$ and $(Y,d)$. Is that correct?

  • 1
    $\begingroup$ well come to MATHEMATICS. Perhaps better ask your questions in three different Questions. $\endgroup$ Aug 29 '15 at 8:34
  • 1
    $\begingroup$ 1 and 3 are correct. $\endgroup$ Aug 29 '15 at 8:39

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