In a metric subspace $S = [0,1]$ of $\mathbb{R}^1$, why is it that every interval of the form $[0,x)$ or $(x,1], x\in (0,1)$, is an open set in $S$?

I understand that if you were to remove either, then the remaining is a closed set; hence both $[0,x)$ or $(x,1]$ are open sets. The definition I'm having difficulty with is the one that involves interior points. WLOG, taking $[0,x)$, if I were to take an open ball centered at some point near $0$ with radius $\epsilon$, then if I wanted to include $0$, the open ball would have to go beyond $0$.

This doesn't quite make sense to me. Could someone please clarify this?

  • 3
    $\begingroup$ The open ball only has to go 'beyond' 0 if there are are actually points 'beyond' 0. However, there are no points 'beyond' 0 in S. $\endgroup$
    – user14972
    Commented Mar 29, 2013 at 22:18

3 Answers 3


In the space $[0,1]$ the open ball of radius $\frac14$ centred at $0$ is by definition


which is clearly just the interval $\left[0,\frac14\right)$. Indeed, for any $r$ satisfying $0<r<1$ the open ball of radius $r$ and centre $0$ is


And if $r>1$, then in $[0,1]$ the open ball of radius $r$ centred at $0$ is


This is all straight from the definition of the open ball in the space $[0,1]$, and it shows that $0$ is in the interior of $\left[0,\frac14\right)$, since there is an open ball centred at $0$ and contained in (in fact equal to) $\left[0,\frac14\right)$.

You can also approach it from the standpoint of the subspace topology on $[0,1]$. A set $U\subseteq[0,1]$ is open in $[0,1]$ in the subspace topology iff there is an open subset $V$ of $\Bbb R$ such that $U=V\cap[0,1]$. Take $U=\left[0,\frac14\right)$, for example: clearly $U=\left(-\frac14,\frac14\right)\cap[0,1]$, and $\left(-\frac14,\frac14\right)$ is certainly open in $\Bbb R$, so $U$ is open in $[0,1]$.

  • $\begingroup$ @Alan: No. What did you think might be one? $\endgroup$ Commented Mar 29, 2013 at 22:52

Since $I=[0,1]$ is being considered as a metric subspace of $\Bbb R$ (with metric $d$, say), then for any $x\in I$ and any $r>0$, letting $B_I(x;r)$ indicate the open $d$-ball in $I$ about $x$ of radius $r$, and $B_{\Bbb R}(x;r)$ be the open $d$-ball of radius $r$ about $x$ in $\Bbb R$, we have $$\begin{align}B_I(x;r) &= \{y\in I:|x-y|<r\}\\ &= I\cap\{y\in\Bbb R:|x-y|<r\}\\ &= I\cap B_{\Bbb R}(x;r).\end{align}$$ Thus, open balls in $I$ around $0$ will either be all of $I$ or will be of form $[0,r)$ for some $0<r<1$. Similarly, open balls in $I$ around $1$ will be all of $I$ or will be of form $(1-r,1]$ for some $0<r<1$.


It's open with respect to the subspace topology. Since you want the entire set to be open, $\left[0,1\right]$ is open with respect to the subspace topology. Then $\left(x,t\right)$ (which is open in $\left(\Bbb{R},\tau_{Euclidean}\right)$), $x\in\left(0,1\right)$, $t \geq 1$, intersected with $\left[0,1\right]$ must be open (as the intersection of a finite number of open sets is open, and $$ \left(x,t\right)\bigcap\left[0,1\right] = (x,1]. $$ Actually, sets of the form $(a,b)$ (where $a$ and $b$ could possibly be infinity) are a basis for the topology on $\Bbb{R}$. Therefore, by definition of the subspace topology, an open set in $\left(\left[0,1\right],\tau\right)$ is given by $U\cap\left[0,1\right]$, where $U$ is open in $\Bbb{R}$, and a basis for this topology is given by sets of the form $(a,b)\cap\left[0,1\right]$.

Note that in the subspace topology on $\left[0,1\right]$, no other points of $\Bbb{R}$ exist, so that given $p\in\left[0,1\right]$, an open set $U$ containing $p$ can be formed by simply taking an open interval $I$ containing $p$ and intersecting it with $\left[0,1\right]$.


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