can i have open sets in closed interval $[0,1]$? I have to show that if $A=[0,1]$ then $(a,1]$ and $[0,b)$ are open in $A$ for $0\le a<1$ and $0<b\le 1$. when i finish my proof a friend tell me that is not true because when $x=1$ in $(a,1]$ i can't find a $r>0$ such that the open ball is complete contain in $(a,1]$ then isn't in $[0,1]$, so, i don't know if true or false that statement because i know in $\Bbb R$ isn't. What do you think?
Edit: My proof consist in find $\epsilon$ but i take $\inf\{1-x,x\}$ for $(a,1]$ and $\inf\{b-x,x\}$ for $[0,b)$. it's the correct choose?
 A: Your friend is wrong and apparently does not understand the subspace topology. A set $U\subseteq A$ is open in $A$ in the subspace topology that $A$ inherits from $\Bbb R$ if and only if there is some ordinary open set $V$ in $\Bbb R$ such that $U=V\cap A$. In your case, for instance, $(a,1]$ is open in $A$ if $0\le a<1$, because $(a,1]=(a,2)\cap A$, and $(a,2)$ is open in the usual topology on $\Bbb R$. Similarly, $[0,b)$ is open in $A$ for $0<b\le 1$ because $[0,b)=(-1,b)\cap A$, and $(-1,b)$ is open in $\Bbb R$.
A: Your friend is wrong. Open sets in $A$ are precisely of the form $U\cap A$ where $U$ is an open set in $\mathbb R$. In particular, $(a,1] = (a,2)\cap [0,1]$ and $[0,b) = (-1,b)\cap[0,1]$.
And more explicitly, $(a,1] = B_A(1,a)$ and $[0,b)=B_A(0,b)$ are open balls in $A$.
A: Your friend is wrong.
An open ball is defined for a radius $r$ and point $k$ then that open ball, $B_r(k)$ is defined to be $\{x \in A| d(x,k) < r\}$.  
And so if $A=[0,1]$ and $0<r < 1$ then $B_r(1) = \{x \in [0,1]| d(x,1) < r\} = (r,1]$.  And that is, by definition, an open ball.
There is nothing in the definition of open ball that says it must be "round" or "protrude equally in all directions".  
I think a difficult concept some students have is that if $A=[0,1]$ is the space, the $[0,1]$ is the entire universe.  There is nothing outside of $[0,1]$.  I may seem like $B_r(1)$ should be $(1-r, 1+r)$ and your friend probably thinks that all the points $(1,1+r)$ or outside $[0,1]$ so that that open ball won't work. But he is wrong.  The points $(1,1+r)$ simply do not exist.  And the open ball is $(1-r,1]$.  That is an open ball.
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Anyhoo.... To prove $(a,1]$ is open: let $x \in (a,1]$.  Then $x > a$ so let $r = x-a$.  Then $B_r(x) = \{y\in [0,1]: d(y,x)< r\} = (x-r, x+r)\cap [0,1] = (a,2x-a)\cap [0,1]$.  If $2x-a \ge 1$ then $(a,2x-a)\cap [0,1] = (a,1]\subset (a,1]$. And if $2x-a < 1$ then $(a, 2x-a) \subset (a,1]$.  
And so $x$ is an interior point even if $x=1$ and $(a,1]$ is open in $[0,1]$.
.....
Or you could simply note that if $w > 1$ then $(a,w)$ is open in $\mathbb R$.  So $(a,w)\cap [0,1]$ is open in $[0,1]$.   And $(a,w)\cap [0,1] = (a,1]$
