# Thread: Real analysis - set theory

1. ## Real analysis - set theory

Hola seņores

I'm just hoping that someone can verify what i write next,

the set $R^{+} = \{ x \in R : x > 0$ }
is closed because the epsilon-neighbourhood around the lower limit, 0, will intersect elements in R-positive

2. Originally Posted by walleye
Hola seņores

I'm just hoping that someone can verify what i write next,

the set $R^{+} = \{ x \in R : x > 0$ }
is closed because the epsilon-neighbourhood around the lower limit, 0, will intersect elements in R-positive

$R^{+}$ is closed if and only if $R^{+}$ contains all its adherent points. $x_{0}$ is an adherent point of $R^{+}$ if for every radius $r>0$, the ball $B_r(x_0)$, that is, { ${x \in R^{+} : |x-x_0|} has a non-empty intersection with $R^{+}$. Is 0 an adherent point and does $R^{+}$ contains 0?

3. Originally Posted by walleye
Hola seņores

I'm just hoping that someone can verify what i write next,

the set $R^{+} = \{ x \in R : x > 0$ }
is closed because the epsilon-neighbourhood around the lower limit, 0, will intersect elements in R-positive

Originally Posted by bram kierkels
$R^{+}$ is closed if and only if $R^{+}$ contains all its adherent points. $x_{0}$ is an adherent point of $R^{+}$ if for every radius $r>0$, the ball $B_r(x_0)$, that is, { ${x \in R^{+} : |x-x_0|} has a non-empty intersection with $R^{+}$. Is 0 an adherent point and does $R^{+}$ contains 0?
Well, I guess that it depends on what topology $\mathbb{R}$ has. If it's the usual topology then the above user has the correct reasoning.