choose the correct option $1)$ $\mathbb{R} \times \{1\}$ $2)$ $\mathbb{R}^2$

Question

let $A = (0,2) \times \{1\}$ . Then A is open in

choose the correct option

$1)$ $\mathbb{R} \times \{1\}$

$2)$ $\mathbb{R}^2$

my attempt : i thinks option $1)$ will be false because $(1- \epsilon , 1 + \epsilon ) \notin \{1\}$

so option $2)$ will be correct

is its true ?

any hints/solution will be appreciated

Answer 1

Hint: prove that $[2,\infty)\times\{1\}$ and $(-\infty,0]$ are closed in $\mathbb{R}\times\{1\}$; similarly, $(-\infty,0])\times\{1\}$.

Further hint: you chose the wrong options.

Answer 2

The correct answer is in fact 1) $\mathbb{R} \times \{1\}$. Every point $a$ in $A$ has, for some $\epsilon >0$, all points within distance $\epsilon$ of $a$ in $\mathbb{R} \times \{1\}$, also in $A$. Every point $a$ in $A$ does NOT have, for any positive $\epsilon$, all points within distance $\epsilon$ of $a$ in $\mathbb{R}^2$, also in $A$.

Indeed:

For each $x \in (0,2)$ and $\epsilon > 0$ the set of points within distance $\epsilon$ of $\langle x,1\rangle$ in $\mathbb{R} \times \{1\}$ is $B_{\epsilon} = \{\langle y,1\rangle; y \in [x-\epsilon, x+\epsilon]\}$ and for each such $x$ there is an $\epsilon >0$ such that $B_{\epsilon} \subset A$.

However for $x \in (0,2)$ and $\epsilon > 0$ the set of points within distance $\epsilon$ of $\langle x,1\rangle$ in $\mathbb{R}^2$ is $B_{\epsilon} = \{\langle y,z\rangle; (y-x)^2+(z-1)^2 \le \epsilon$ and for each such $x$ there is no such $\epsilon >0$ such that $B_{\epsilon} \subset A$.

Answer 3

Hint: Since both are metric spaces (prove it!), every point $x$ of an open set $U$ must have an open ball $B(x,\varepsilon)$ for some $\varepsilon > 0$ such that $B(x,\varepsilon) \subset U$.

How are the open balls in each of these spaces? In which spaces does this property holds true for $A$?

Answer 4

In questions like these, it is important to understand the topology of the metric spaces you're working with. In the case of products of metric spaces, the question usually refers to the product topology - a set $A$ is open if and only if it is a union of products of open balls from the spaces that define the product.

As you can see, under the product topology, $A$ actually is open in $\mathbb{R}\times ${$1$}, since it can be displayed as $A=(0,2)\times ${$1$}, which is a product of two open balls (each is open in its own corresponding metric space).

On the contrary, $A$ is not open in $\mathbb{R}^2$, since it is not a union of open balls in $\mathbb{R}^2$ (there is no open ball around any point of the set).

Answer 5

Perhaps an overkill:

1) $\mathbb{R}×${$1$} $\subset \mathbb{R^2}$.

$A \subset \mathbb{R}×${$1$} is open in $\mathbb{R}×${$1$}

$ \iff$ there is a set $O$, open in $\mathbb{R^2},$ s.t.

$A= O \cap \mathbb{R}×${$1$}.

Choose $O = (0,2)×(0,2)$ which is open in $\mathbb{R^2}$.

$A= O \cap \mathbb{R}×${$1$}, and we are done.

2) Let $a =(1,1) \in A$.

$B_r(a) \not \subset A$ for $r>0$, since

$x = (1,1+r/2) \in B_r(a)$ but $x \not \in A$ for $r>0$.