How to attempt the following question?

Question: Let $Y=[0,1]$, considered as a subspace of $\mathbb{R}$ with the topology $\tau=\{ I \subseteq \mathbb{R} \mid I\text{ an open interval}\}.$ Find the subspace topology.

Here are my steps:

1. Let $I = (a,b)$. Let $\underline{T}={Y\cap U\mid U\tau T}$.
• $(a ,b) \cap Y= (a, b)$ if $a, b \in Y$.
• $(a, b)\cap Y=(0, b]$ if only $b \in Y$.
• $(a ,b)\cap Y= (a,1]$ if only $a \in Y$.
• $(a, b)\cap Y= Y$ or $\emptyset$ if neither $a$ nor $b$ are in $Y$. Then, $\underline{T}= \{ \emptyset, [0,b), (a,1], Y\}$.

I am not sure whether I can include $(a, b)$ in $\underline{T}$!

$\underline{T}=\{ \emptyset, [0,b), (a,1], Y\}$

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Once you receive a satisfactory answer to your question, you should "accept" it by clicking on the check mark on the left side of an answer, near the vote up/vote down buttons. –  Tyler Apr 8 '11 at 14:34
You can! Your reasoning is entirely right. You even have written on one line, $(a,b) \cap Y = (a,b)$ if $(a,b) \in Y$. Without this you have not specified the topology completely. (I do think you have specified a correct subbasis for the topology) –  Jay Kopper Apr 8 '11 at 14:42
Please take some time to format your posts. There is no need to put three or four spaces between every word, and you should spend a bit of time learning some of the basic mark-up. This post was a mess. –  Arturo Magidin Apr 8 '11 at 15:06
$\tau$ is not a topology. –  Arturo Magidin Apr 8 '11 at 15:10
@leopard: First, state the problem correctly. Then, do some thinking. –  Arturo Magidin Apr 8 '11 at 15:18

$\tau$ is not a topology on $\mathbb{R}$: it is not closed under arbitrary unions (the union of two disjoint open intervals is not an open interval). So the entire problem starts on the wrong foot.

Added. It's possible that $\tau$ is meant to be the topology generated by the open intervals.

Your description of $(a,b)\cap Y$ when $a\notin Y$ and $b\in Y$ is incorrect. Presumably, you meant $(a,b)\cap Y = [0,b)$, not $(0,b]$.

Your description of $\underline{T}$ is incorrect; first, you shouldn't just be considering a single interval $(a,b)$; you should be considering all intervals. Certainly $\underline{T}$ contains more than just four things in it.

What makes you doubt whether you have subsets of the form $(a,b)$ in $\tau\cap Y$? If $0\leq a\lt b\leq 1$, then $(a,b)\cap Y = (a,b)$, so $(a,b)\in\tau\cap Y$.

Added. So if you want to consider the topology generated by the open intervals, first you should give an appropriate/correct description of the intersection with $Y$, and then check to see that if you let $\tau$ be the topology generated by the open intervals on $\mathbb{R}$, then the induced topology on $Y$ is equal to the topology generated by the intersections of open intervals with $Y$.

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thanks, i have to inform my subject coordinator –  leopard Apr 8 '11 at 15:50
@leopard: You also need to find where the shift key in your keyboard is hiding... –  Arturo Magidin Apr 8 '11 at 16:00