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

choose the correct option

$1)$ $mathbbR times 1$

$2)$ $mathbbR^2$

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

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

Indeed:

For each $x in (0,2)$ and $epsilon > 0$ the set of points within distance $epsilon$ of $langle x,1rangle$ in $mathbbR times 1$ is $B_epsilon = langle y,1rangle; y in [x-epsilon, x+epsilon]$ 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 $langle x,1rangle$ in $mathbbR^2$ is $B_epsilon = {langle y,zrangle; (y-x)^2+(z-1)^2 le epsilon$ and for each such $x$ there is no such $epsilon >0$ such that $B_epsilon subset A$.

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 $mathbbRtimes $$1$, since it can be displayed as $A=(0,2)times $$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 $mathbbR^2$, since it is not a union of open balls in $mathbbR^2$ (there is no open ball around any point of the set).

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 $A$?

Hint: prove that $[2,infty)times1$ and $(-infty,0]$ are closed in $mathbbRtimes1$; similarly, $(-infty,0])times1$.

Further hint: you chose the wrong options.

Perhaps an overkill:

1) $mathbbR×$$1$ $subset mathbbR^2$.

$A subset mathbbR×$$1$ is open in $mathbbR×$$1$

$ iff$ there is a set $O$, open in $mathbbR^2,$ s.t.

$A= O cap mathbbR×$$1$.

Choose $O = (0,2)×(0,2)$ which is open in $mathbbR^2$.

$A= O cap mathbbR×$$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 $r>0$.