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On the metric subspace $S = [0,1]$ of the Euclidean space $\mathbb R^1 $, every interval of the form $A = [0,a)$ or $(a, 1]$ where $0<a<1$ is open set in S. These sets are not open in $\mathbb R^1$

Here's what I attempted to show that $A$ is open in $S$. I have no idea how it is not open in $\mathbb R^1$.

Let $M = \mathbb R^1$, $x \in A = [0,a)$ .
If $x = 0$, $ r \leq \min \{a, 1-a\}, \\\ B_S(0; r) = B_M(0,r)\cap[0,1] = (-r, r) \cap[0,1] = [0,r) \subseteq A $

If $x \neq 0, r \leq \min \{x,|a-x|, 1-x\}, \\B_S(x; r) = B_M(x,r)\cap[0,1] = (x-r, x+r) \cap[0,1] \subset (0,x) \text{ or } (x, 1) \subset A$

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Note that if $x=0$, you are not able to consider the case $x\in (a,1]$, so you may just take $r=a$. Also if $x\neq 0$, note that $|a-x|=|x-a|$. So this shortens also a little bit. And how is it possible that $(x-r,x+r)\subset (0,x)$ or $(x,1)$? – T. Eskin Dec 29 '12 at 21:40
@ThomasE. thanks for correction!! – Santosh Linkha Dec 29 '12 at 21:42
up vote 4 down vote accepted

Hint: To show $(x,1]$ not open in $\mathbb R$ simply show that there is no open neighborhood of $1$ included in the half-closed interval.

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The open sets in $[0,1]$ using the subspace topology are those of the form $U \cap [0,1]$ where $U$ is open in $\mathbb{R}$.

$[0,x)$ is open because, for example, it can be written as $(-1,x) \cap [0,1]$ and $(x,1]$ is open because it can be written as $(x,2) \cap [0,1]$.

$[0,x)$ is neither open in $\mathbb{R}$ because every open set containing $0$ contains points in $(-\infty,0)$, and is not closed because $x \notin [0,x)$, but $(x-\frac{1}{n}) \to x$, and $x-\frac{1}{n} \in [0,x)$ forall $n$ sufficiently large.

Similarly $(x,1]$ is not open. (Explicitly, the map $y \mapsto 1-(\frac{1-x}{x}) y$ is a homeomorphism of $[0,x)$ onto $(x,1]$.)

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Recall that open sets in $[0,1]$ can be characterized as all intersections of open sets in $\mathbb R$ with $[0,1]$. If you don't know this result, consult your textbook. Can you find an open set $U$ so that $U \cap [0,1]$ is $[0,x)$? What about $(x,1]$?

To show these half-open intervals are not open in $\mathbb R$, recall what is means to be open in $\mathbb R$. Find a point so that no neighborhood of the point is contained in the set. For example, if you have $[0,x)$, look at neighborhoods of $0$.

(This answer assumes you are using the subspace topology on $[0,1]$. If you don't know what this means, ignore this parenthetical comment.)

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