# Math Help - Usual topology

1. ## Usual topology

My professor gave us a question in exam

Let $A=[-2,2]$ and consider the subspace $(A,T_A)$ of the space $(R,T_u)$ .Take

$B=\{x : \mid x \mid <1\}-(\{\frac{1}{n} : n\in N \})$

find a $T_u$ open subset $W\subseteq R$ such that
$B=A\cap W$ . Conclude that B is $T_A$ open

I answered W=B since B is open in $T_u$ and $B\subseteq A$ so $A\cap W = A\cap B=B$

but after he finished the correction of the exam and he want to gave us it he said that the question is deleted since he forgot to set a side {0} from B so B present like this

$B=\{x : \mid x \mid <1\}-(\{\frac{1}{n} : n\in N \}\cup \{0\})$

and he said that B in the first form is not $T_u$ open !! and I ask him why?? he answered me but I did not got it and I shamed to ask him again .

so can you explain why B in the first form is not open although it is an arbitrary union of open sets in $T_u$ and the arbitrary unoin of open set is open .

and we can write B like this $B=(-1,\frac{1}{2})\cup (\frac{1}{2},\frac{1}{3})\cup (\frac{1}{3},\frac{1}{4})\cup (\frac{1}{4},\frac{1}{5})\cup.....$

Thanks

2. Originally Posted by Amer

and we can write B like this $B=(-1,\frac{1}{2})\cup (\frac{1}{2},\frac{1}{3})\cup (\frac{1}{3},\frac{1}{4})\cup (\frac{1}{4},\frac{1}{5})\cup.....$

Thanks
See I think part of the propblem is your fractions are screwed up, that first set in the union is going to contain everything else in your union, so all that is unnecessary, and i think we can all agree $B \not = (-1, \frac{1}{2})$. Recall $\frac{1}{n}\rightarrow 0$.
I see what you are trying to do though and see the confusion.

By definition, a set B is open iff for every x in B, there is an open set containing x and contained in B.

But consider that point 0 that is in B. There is no open interval (these are the basis elements for the real line in the usual topology) which contains 0 and does not intersect the sequence $\{\frac{1}{n}\}$ (in fact there are an infinite number of points of this intersection for every open set containing 0 by definition of $\frac{1}{n}\rightarrow 0$).

If instead 0 were removed like the teachers correction states (call it B') what you were trying to do is spot on.
$B'=(-1,0)\bigcup_{n=1}^\infty(\frac{1}{n+1},\frac{1}{n} )$
but the way it stands as he originally stated, 0 is in B, but not in B'.

3. Originally Posted by Gamma
See I think part of the propblem is your fractions are screwed up, that first set in the union is going to contain everything else in your union, so all that is unnecessary, and i think we can all agree $B \not = (-1, \frac{1}{2})$. Recall $\frac{1}{n}\rightarrow 0$.
I see what you are trying to do though and see the confusion.

By definition, a set B is open iff for every x in B, there is an open set containing x and contained in B.

But consider that point 0 that is in B. There is no open interval (these are the basis elements for the real line in the usual topology) which contains 0 and does not intersect the sequence $\{\frac{1}{n}\}$ (in fact there are an infinite number of points of this intersection for every open set containing 0 by definition of $\frac{1}{n}\rightarrow 0$).

If instead 0 were removed like the teachers correction states (call it B') what you were trying to do is spot on.
$B'=(-1,0)\bigcup_{n=1}^\infty(\frac{1}{n+1},\frac{1}{n} )$
but the way it stands as he originally stated, 0 is in B, but not in B'.
thanks very much I understand it .