# Complement and open set on topological spaces

Question: Let $Y=\left [ -1,1 \right ]$ be a subspace of $\mathbb{R}$ so the subspace property holds. Is $\left \{ x:\frac{1}{2}\leq \left | x \right |<1 \right \}$ an open set?

From the definition of complement,

On a metric space, a proper subset V of X is an open set wrt X if its complement $X\setminus V=\left \{ x \in X : x \notin V \right \}$

Here, $Y\setminus A=\left \{ y \in Y | y \notin A \right \}=\left ( \frac{-1}{2},\frac{1}{2} \right )\cup \left \{ -1 \right \}\cup \left \{ 1 \right \}$

Now, to speak about an open set we have to talk about open balls. I do not have good exposure to elementary real analysis so I'm not exactly sure how can we speak about open ball in this question.

Any help is appreciated.

Edit:

Let $\bar{y}$ be an element in the complement $Y\setminus A=\left ( -1,\frac{-1}{2} \right ] \cup [\frac{1}{2},1 )$ The pertinent is this:

Is there any element $\bar{y} \in Y\setminus A$ st $d\left ( \bar{y},y \right )<\epsilon$ but $B_{\epsilon}\left ( \bar{y} \right )\nsubseteq Y\setminus A$?

• You mean $Y$ is a subset rather than a subspace, don't you? – user190080 Aug 27 '16 at 8:59
• Changes made to OP – Mathematicing Aug 27 '16 at 9:02
• I don't understand the solution by the poster below. I've made an edit to the OP. Would someone assist me? – Mathematicing Aug 27 '16 at 10:56

Well, you say you are operating in a metric space $(\mathbb R,d(x,y))$ and further you're given the subset $$\left \{ x:\frac{1}{2}\leq \left | x \right |<1 \right \}=(-1,-\frac 1 2]\cup[\frac1 2,1)=:X\subset Y\subset \mathbb R$$ A set is called an open set if for any point $x\in X$ there exists an $\varepsilon>0$ such that for any given point $y\in Y (\text{or }\mathbb R)$ with $d(x,y)<\varepsilon$, $y$ also belongs to $X$.
Now let's take for example $\displaystyle\frac 1 2\in X$ and any $\varepsilon>0$, then $$\left\{y:y\in Y, d(\frac 1 2,y)<\varepsilon\right\}\not\subset X$$ and therefore your set $X$ cannot be an open set in $Y$ (or especially in $\mathbb R$ for this matter).
• I show that there is a point in $X$, $\frac 12$ for which there exists no neighborhood $d(\frac 12,y)$ which is a subset of $X$ - this means, that $X$ is not open. Alternatively you could also show, that the complement is not closed, which I guess was your first idea to do? – user190080 Aug 27 '16 at 9:43