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I am new to topology and I want to understand this example from wikipedia about subspace topology.

The example is:

Let $S = [0, 1)$ be a subspace of the real line $R$. Then $[0, 1/2)$ is open in $S$ but not in $R$. Likewise $[½, 1)$ is closed in $S$ but not in $R$. $S$ is both open and closed as a subset of itself but not as a subset of $R$.

I cannot understand the part where it says:

Likewise $[½, 1)$ is closed in $S$ but not in $R$.

Does the phrase mean the it is closed in $S$ but open in $R$? I hope someone can help me understand it better.

Thank you.

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  • $\begingroup$ No, the phrase does not imply that $[1/2,1)$ is open in $\mathbb{R}$. Just to make sure that you understand, there are sets which are neither open nor closed in $\mathbb{R}$. Hence, a set which is not closed does not have to be open. And vice versa. $\endgroup$
    – shalop
    Mar 22 '15 at 8:18
  • $\begingroup$ I understand now. Thanks a lot. $\endgroup$
    – Nikki Mino
    Mar 22 '15 at 8:30
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Usee the fact that

Let $Y$ be a subspace of a topological space $X$. Then a set $A$ is closed in $Y$ if and only if it equals the intersection of a closed set of $X$ with $Y$.

Clearly $[\frac{1}{2},1)=[\frac{1}{2},1]\cap S$ is the intersection of a closed subset of $\Bbb R$ with $S$ and so $[\frac{1}{2},1)$ is closed in $S$. But $[\frac{1}{2},1)$ is not closed in $\Bbb R$ as $1$ is the limit point of that set which do not belong to that set (or, the complement $\Bbb R\setminus [\frac{1}{2},1)=(-\infty,\frac{1}{2})\cup[1,\infty)$ is not open in $\Bbb R$).

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  • $\begingroup$ Thank you very much for your help $\endgroup$
    – Nikki Mino
    Mar 22 '15 at 8:29
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A set is said to be open if its complement is closed and vice versa. So if we consider $S\setminus [\dfrac{1}{2},1)$ then

$S\setminus [\dfrac{1}{2},1)=[0,\dfrac{1}{2})$ which is open in $S$. Hence $[\dfrac{1}{2},1)$ is closed in $S$. But this set is not closed in $\mathbb{R}$ because $\mathbb{R}\setminus[\dfrac{1}{2},1)=(-\infty,\dfrac{1}{2})\cup[1,\infty)$.

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Consider $S\setminus [1/2,1)$. $S\setminus [1/2,1)=[0,1/2)$ is open in $S$. So $[1/2,1)$ is closed in $S$ (since complement of $[1/2,1)$ in $S$ is open in $S$). But $[1/2,1)$ is not closed in $\mathbb{R}$ since $1$ is a limit point of $[1/2,1)$ and $1\notin [1/2,1)$.

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