If I had a set such as $[1,\infty)$ or $(-1,\infty)$ how could I prove whether the sets are open or closed. I know the $\epsilon$ definition of openness and the fact that a set is closed if it contains all of its limit points. I want to start by fixing an arbitrary limit point in the set, but I need help figuring out what to do next.
Determine whether a set is open or closed in R
general-topologyreal-analysis
Related Solutions
Suppose $C^c$ is open. Suppose $x $ is a limit point of $C$. If $x \notin C$ then there is some open $V$ such that $x \in V \subset C^c$, which contradicts $x$ being a limit point.
Suppose $C$ contains all its limit points. Suppose $x \notin C$. Since $x$ is not a limit point, there is some open $V$ such that $x \in V$ and $V$ does not intersect $C$. Hence $C^c$ is open.
(1).Your answers are correct and your reasoning has no flaws.
(2).The usual notation for an open ball, of radius $e$, centered at $x$, in a metric space $(X,d)$, is $B_d(x,e)$. When only one metric ($d$) is under consideration the subscript "$d$" is often omitted : $B(x,e)$. (Caution: In general we can have $B_d(x_1,e_1)=B_d(x_2,e_2)$ with $x_1\ne x_2$ or $e_1\ne e_2 $ or both.)
(3).Your def'n of relatively open (your last paragraph) is flawed: For metric space $(X,d)$ and $A\subset D\subset X,$ the set $A$ is relatively open in $D$ iff $$\forall x\in A \;\exists e>0\;(B_d(x,e)\cap D\subset A).$$
(4). A general approach: Let $T$ be a topology (the set of open sets) on a set $X.$ (Look up, if necessary, the general def'n of a topology). Closed sets are defined as the complements of open sets.
(5).In a metric space, the topology defined by (generated by ) the metric is: A set is open iff it is the union of a set of open balls. We can deduce that in a metric space a set is closed iff it contains all its metric limit points.
(6).Let $T$ be any topology on a set $X.$ Let $Y \subset X.$ The subspace topology on $Y,$ with respect to the topology $T,$ is $$\{t\cap Y: t\in T\}.$$ From this it is immediate that if $Z\subset Y$ then $Z$ is relatively open in $Y $ iff $Z=t\cap Y$ for some $t$ that is open in $X.$ (In particular if $Z\subset Y$ and $Z$ is open in $X$ then $Z$ is relatively open in $Y$.)
Deduce that for a metric space, this def'n of relatively open is equivalent to the def'n of relatively open in (3).
(7). For $S\subset X,$ the closure of $S$ in $X,$ sometimes denoted $Cl_X(S)$, is defined as the intersection of all closed subsets of $X$ that have $S$ as a subset. Equivalently, $X$ \ $Cl_X(S)$ is the union of all open subsets of $X$ that are disjoint from $S$.
And $S$ is closed in $X$ iff $S=Cl_X(S)$.
For $S\subset Y\subset X,$ the closure of $S$ in $Y$, denoted $Cl_Y(S)$ is the intersection of all subsets of $Y$ that are relatively closed in $Y$ and contain S as a subset. And if $S \subset Y$ then $S$ is closed in $Y$ iff $S=Cl_Y(S).$
An important, useful deduction is $$Cl_Y(S)=Y\cap Cl_X(S) \;\text {for all }\; S\subset Y.$$ In particular, if $Y\supset S=Cl_X(S)$ then $S$ is relatively closed in $Y$.
(8). Finally, with $X=\mathbb R$ and $Y=[0,2)$ we have:
(i)...$[0,1]$ is closed in $X$ so it is closed in $Y$. If $t$ is an open subset of $X$ and $t\supset [0,1]$ then $t\supset [1,1+r)$ for some $r\in (0,1).$ So $t\cap Y\supset [1,1+r)$ so $t\cap Y\ne [0,1].$ So $[0,1]$ is not open in $Y.$
Or we may observe that $Cl_Y(Y$ \ $[0,1])=Y\cap Cl_X((1,2))=Y\cap [1,2]=[1,2)\ne Y$ \ $[0,1]$, so therefore $Y$ \ $[0,1]$ is not closed in $Y$.
(ii)...The same argument that shows that $[0,1]$ is not open in $Y$ will also show that $(0,1]$ is not open in $Y.$ And $Cl_Y((0,1])=Y\cap Cl_X((0,1]=Y\cap [0,1]=[0,1]\ne (0,1]$ so $(0,1]$ is not closed in $Y.$
(iii)...$(-1,1)$ is open in $X$ so $(-1,1)\cap Y=[0,1)$ is open in $Y$. And $Cl_Y([0,1))=Y\cap Cl_X([0,1))=Y\cap [0,1]=[0,1]\ne [0,1)$ so $[0,1)$ is not closed in $Y.$
(iv). Since $[0,1)$ is open but not closed in $Y ,$ its complement in $Y,$ which is $[1,2),$ is closed but not open in $Y$.
Best Answer
There are two ways to go about this. First, we should decide whether each of these sets are open or closed. Let's start with $[1,\infty)$. Here, it's best to look at the complement $$ \mathbb{R} \setminus [1,\infty) = (-\infty, 1). $$ Note that the above is an open interval, and is thus open in $\mathbb{R}$. Consequently, $[1,\infty)$ is closed. Alternatively, one can use a sequential argument. For $[1,\infty)$ to be closed, we must show that every convergent sequence $(x_n)$ of points in $[1,\infty)$ has its limit in $[1,\infty)$. More precisely, assume that $(x_n)$ is a sequence in $[1,\infty)$ converging to some $x \in \mathbb{R}$. Since $x_n \geq 1$ for all $n$, it is clear that $x \geq 1$ as well. That is, $$ x = \lim{x_n} \in [1,\infty). $$ From this, we deduce that $[1,\infty)$ is closed.
Now, we treat the second set $(-1,\infty)$. Again, this is an open interval and is thus (by definition) open.