On the real line $\mathbb{R}$, if one uses the usual metric $d(x,y) = |x−y|$, then the set ${\{1\}}$ is not open, however if instead one uses the discrete metric $d_{disc}$, then ${\{1\}}$ is now an open set.
I don't quite understand the reasoning behind the above statement. Also would it be closed on $\mathbb{Q},\mathbb{Z}?$
I know that a space is defined as closed, if it contains its boundary, but for the set ${\{1\}}$ I am not sure which is the boundry?
On
If a set $X$ is equipped with the discrete metric then all singletons are open because $\{p\} = \{x\in X : d(x,p) < 1\}$.
If we equip $\Bbb R$ with $d(x,y) = |x-y|$, then $\{1\}$ is closed because its complement is open (any point in the complement is positive distance away from $1$). Also bear in mind that "open" is not the same as "not closed" because a set can be both open and closed.
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A set is closed iff its complement is open. Consider $S=\mathbb{R} - \{1\}$. Then if $x\in S$ it’s not hard to find a neighborhood contained in $S$ and containing $x$.
On
The subset $A$ of a metric space $X$ is referred to as open if for any $x\in A$ we have an $\epsilon>0$ such that $B_\epsilon(x)\subset A$ either where $$B_\epsilon(x)=\{a:d(a,x)<\epsilon\}$$ By using discrete metric such thing happens for every $\epsilon<1$ because:$$B_\epsilon(x)=\{a:d(a,x)<\epsilon<1\}=\{a:d(a,x)=0\}=\{a\}\subset A$$which implies on openness. Also it is closed so a more likely phrase is "Clopen" which are both open and closed. Also since it is clopen in $\Bbb R$ so is in $\Bbb Q$ or $\Bbb Z$ according to "Inheritance principle".
For this problem, a better (and equivalent) definition of closed is: $A$ is closed if and only if $\Bbb{R}\setminus A$ is open. So, in the usual topology, pick any real number $x$ not in the set $\{ 1 \}$. Clearly there is some $\varepsilon > 0$ such that $(x-\varepsilon, x+\varepsilon)$ does not include $1$. Thus the complement of $\{ 1 \}$ is open, so $\{ 1 \}$ is closed.
If you would rather use the "contains its boundary" definition, then consider this: $1$ is in the boundary with the usual metric because every open ball around $1$ will contain both the point $1$ and points not in $\{ 1 \}$. So, $1$ is in the boundary, and it is in the set, so the set contains its boundary.
In the discrete metric, $\{ 1 \}$ is the ball with radius $1/2$ centered at 1, so it contains an open ball around all of its points, and is thus open.