Doubt about sets open in metric subspaces need not be open in the respective metric space.

Question

In the example given the set [0,a),(a,1] is open on set S=[0,1] but closed on R. As there are no examples provided to illustrate and directly proceeded to other theorems, i am trying to decipher. Cant we choose radius 'r' finitely large such that we can find an open ball for which d(x,y)<r . or does it happen that the given set will have points which are not interior to the set. What exactly is happening on R which is not happening on S, that the set is not open on R. Hope my doubt is understandable. Thank You.

[Theorem][1]

[1]: https://i.stack.imgur.com/c928H.png**strong text**

Did you mean $\color{red}[0,a)$ with $0<a<1$? That is neither open nor closed in $\mathbb R$. It is open in $S=[0,1]$ because it is the intersection of $S$ and an open subset of $\Bbb R$, say $(-a,a)$ — J. W. Tanner, Nov 23 '20 at 14:54
$[0,a)$ is not open in $\mathbb R$ because it does not contain a neighborhood of $0$ — J. W. Tanner, Nov 23 '20 at 15:02
@J.W.Tanner does containing neighborhood mean existence of open ball? — aravindd7, Nov 23 '20 at 15:06
Yes, in $\Bbb R$, there is no open ball around $0$ in $[0,a)$ — J. W. Tanner, Nov 23 '20 at 15:19

score 1 · Answer 1 · answered Nov 23 '20 at 23:25

A topological space is a pair $(X, T_X)$ where $X$ is any set and $T_X$ is a collection (a set) of some or all of the subsets of $X,$ subject to:

$\phi$ (the empty set) and $X$ belong to $T_X.$
If $C$ is any set of members of $T_X$ then the common union $\cup C=\cup_{t\in C}\,t $ belongs to $T_X.$
If $D$ is any $finite$ set of members of $X$ then the common intersection $\cap D=\cap_{d\in D}\,d$ belongs to $T_X.$

$T_X$ is called a topology on $X.$ The members of $T_X$ are called the open sets, but this is meaningful only when talking about a particular $T_X.$ And the complements, in $X,$ of the members of $T_X$, are called the closed sets. Again, this makes sense only when talking about a particular $T_X.$ It is common to say "the space $X$", especially when it is understood from the context (that is, from other things that were already said), which $T_X$ is being considered.

When $(X,T_X)$ is a topological space and $S\subseteq X$ we can define a topological space $(S,T_S)$ where $T_S=\{t\cap S:t\in T_X\}.$ This $T_S$ is called the subspace topology on $S$ with respect to $T_X.$ It is common to say "the subspace $Y$".

If $T_{\Bbb R}$ is the "usual" or " standard" topology on $\Bbb R,$ then any $C\subset \Bbb R$ belongs to $T_{\Bbb R}$ iff (if & only if) for every $x\in C$ there exists $r_x>0$ such that $(-r_x+x,r_x+x)\subseteq C.$ With respect to this topology, if $S=[0,1]$ then $[0,a)\cup (a,1]$ belongs to the subspace topology on $S$ because the set $t=(-1,a)\cup (a,2)$ belongs to $T_{\Bbb R}$ and $[0,a)\cup (a,1]=t\cap S.$

Doubt about sets open in metric subspaces need not be open in the respective metric space.

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