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.

Thanks in advance.


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 $B_{\epsilon}\left ( \bar{y} \right )\nsubseteq Y\setminus A$?

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

  • $\begingroup$ I don't understand. You want to show that every element in the complement satisfies the open set definition and the arrive at a conclusion. Why you taking an element in X?. All you're really doing is to show whether or not X is an open set $\endgroup$ – Mathematicing Aug 27 '16 at 9:38
  • $\begingroup$ 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? $\endgroup$ – user190080 Aug 27 '16 at 9:43
  • $\begingroup$ What is that second bracket? Is it an open ball the set of elements in Y? $\endgroup$ – Mathematicing Aug 27 '16 at 9:51
  • $\begingroup$ What second bracket do you mean? $\endgroup$ – user190080 Aug 27 '16 at 9:57
  • $\begingroup$ The second expression $\endgroup$ – Mathematicing Aug 27 '16 at 10:08

Y=[-1,1] which is closed subset of reals... Let X ={x|1/2<=|x|<1} or we can say X=(-1,-1/2] union [1/2,1) It is neither open nor closed.... Complement of X in Y is the same as given in the question....i.e. (-1/2,1/2)union {-1} union{1}…… Now complement is not open as it is not the nbd. of -1 and 1.....or we can say -1 and 1 are not interior points of complement of X.....hence X cannot be closed.... X is not open as....-1/2 and 1/2 are not interior points of X..... Hence X is neither open nor closed...... (For interior points there should exist at least one nbd. of radius epsilon which is contained in that set.....)


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