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

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

3. Originally Posted by walleye
Hola seņores

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

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

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