# Is it true that open intervals on the real line and the real line is closed in $\mathbb{R}^2$

I am trying to construct a closed connected set in $\mathbb{R}^2$ and I vaguely remember that an open interval is considered closed in $\mathbb{R}^2$ but I lack a satisfactory explanation.

Claim: open intervals on the real line and real line are closed in $\mathbb{R}^2$

The argument was that since each point is surrounded by a ball, and that ball necessarily contains points not in the open intervals, therefore the intervals are not open. But this is not good because it does not imply that they are closed.

Can someone please verify whether open intervals are considered closed in $\mathbb{R}^2$?

• How could that be? The line segment from $(0,0)$ to $(1,0)$, without endpoints, doesn't contain its two limit points, namely, the endpoints. Jan 28 '16 at 4:10
• Only closed subintervals of $\mathbb R$ are closed in the plane. If you want a closed connected subset of $\mathbb R^2,$ why not use $\mathbb R^2?$
– zhw.
Jan 28 '16 at 4:10
• Well they are not closed. Any open interval $(a,b)$ will not be closed as it doesn't contain its limit points $a$ and $b$
– R_D
Jan 28 '16 at 4:10

An open interval $(a,b)$ is not closed in $\mathbb{R}^2$ because there are sequences of elements in $(a,b)$ converging in $\mathbb{R}^2$ whose limits aren't in $(a,b)$. Just take a sequence approach $a$ in $(a,b)$. In general this argument shows that things that aren't closed in $\mathbb{R}$ aren't going to all of a sudden become closed in $\mathbb{R}^2$.
Open intervals on the real line in $\mathbb{R}^2$ are not closed because their endpoints are the limits of sequences in those intervals and yet the endpoints do not belong to the intervals.
• Note $(-\infty,\infty)$ is closed in $\mathbb R^2.$
• Apologies. I didn't think of $(-\infty, \infty)$ (the entire real line) as an open interval, but maybe that is something I've forgotten, or there are contexts where it is so called. Jan 28 '16 at 4:20
Open intervals in $\mathbb R$ are not closed when included in $\mathbb R^2$. To see this, we have to go back to the definition of closed. A set $A$ is closed if its compliment $A^C$ is open. The set $A^C$ is open if for every $x\in A^C$, there is a number $\epsilon$ so that the set $B_\epsilon(x)=\{y:|x-y|<\epsilon\}\subset A^C$. Let $A$ be the open "interval" contained in $\mathbb R^2$, $\{(x,0):a<x<b\}$. Then the point $(a,0)$ is in $A^C$, but there is $\epsilon$ small enough so that $B_\epsilon(x)\subset A^C$, so $A^C$ is not open, which means that $A$ is not closed. As it happens, $A$ is neither closed nor open.