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The Topology on the real line R generated by left open right closed interval $(a,b]$ is Choose the correct...
TheTopology on the real line R generated by left open right closed interval $(a,b]$ is
Choose the correct option
$a)$ strictly coarser than the usual topology
$b)$ strictly finer than the usual topology
$c)$ NOT comparable with usual Topology
$d)$ same as the usual Topology
My attempts : I thinks option $b)$ will correct because
$$
(a,b) = cup_{nge 1} (a ,b+frac{epsilon}{n}]
$$
where $epsilon < frac{b-a}{2}$
This implies $(0,1]$ is not open in the usual topology.
is Its correct??
general-topology
asked Nov 20 '18 at 21:03
jasmine
1,536416
add a comment |
TheTopology on the real line R generated by left open right closed interval $(a,b]$ is
Choose the correct option
$a)$ strictly coarser than the usual topology
$b)$ strictly finer than the usual topology
$c)$ NOT comparable with usual Topology
$d)$ same as the usual Topology
My attempts : I thinks option $b)$ will correct because
$$
(a,b) = cup_{nge 1} (a ,b+frac{epsilon}{n}]
$$
where $epsilon < frac{b-a}{2}$
This implies $(0,1]$ is not open in the usual topology.
is Its correct??
general-topology
asked Nov 20 '18 at 21:03
jasmine
1,536416
add a comment |
TheTopology on the real line R generated by left open right closed interval $(a,b]$ is
Choose the correct option
$a)$ strictly coarser than the usual topology
$b)$ strictly finer than the usual topology
$c)$ NOT comparable with usual Topology
$d)$ same as the usual Topology
My attempts : I thinks option $b)$ will correct because
$$
(a,b) = cup_{nge 1} (a ,b+frac{epsilon}{n}]
$$
where $epsilon < frac{b-a}{2}$
This implies $(0,1]$ is not open in the usual topology.
is Its correct??
general-topology
asked Nov 20 '18 at 21:03
jasmine
1,536416
TheTopology on the real line R generated by left open right closed interval $(a,b]$ is
Choose the correct option
$a)$ strictly coarser than the usual topology
$b)$ strictly finer than the usual topology
$c)$ NOT comparable with usual Topology
$d)$ same as the usual Topology
My attempts : I thinks option $b)$ will correct because
$$
(a,b) = cup_{nge 1} (a ,b+frac{epsilon}{n}]
$$
where $epsilon < frac{b-a}{2}$
This implies $(0,1]$ is not open in the usual topology.
is Its correct??
general-topology
general-topology
asked Nov 20 '18 at 21:03
jasmine
1,536416
asked Nov 20 '18 at 21:03
jasmine
1,536416
asked Nov 20 '18 at 21:03
jasmine
1,536416
asked Nov 20 '18 at 21:03
jasmine
1,536416
asked Nov 20 '18 at 21:03
jasmine
1,536416
1,536416
add a comment |
add a comment |
1 Answer
1
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votes
Yes, it is correct. How about $[a,b)$ ?Is it the same ?also there are many typologies that finer and coarser than the usual topology
answered Nov 20 '18 at 21:27
gb2017
944
thanks u @gb2017
– jasmine
Nov 20 '18 at 21:30
add a comment |
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1 Answer
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1 Answer
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active
oldest
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active
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votes
Yes, it is correct. How about $[a,b)$ ?Is it the same ?also there are many typologies that finer and coarser than the usual topology
answered Nov 20 '18 at 21:27
gb2017
944
thanks u @gb2017
– jasmine
Nov 20 '18 at 21:30
add a comment |
Yes, it is correct. How about $[a,b)$ ?Is it the same ?also there are many typologies that finer and coarser than the usual topology
answered Nov 20 '18 at 21:27
gb2017
944
thanks u @gb2017
– jasmine
Nov 20 '18 at 21:30
add a comment |
Yes, it is correct. How about $[a,b)$ ?Is it the same ?also there are many typologies that finer and coarser than the usual topology
answered Nov 20 '18 at 21:27
gb2017
944
Yes, it is correct. How about $[a,b)$ ?Is it the same ?also there are many typologies that finer and coarser than the usual topology
answered Nov 20 '18 at 21:27
gb2017
944
answered Nov 20 '18 at 21:27
gb2017
944
answered Nov 20 '18 at 21:27
gb2017
944
answered Nov 20 '18 at 21:27
gb2017
944
944
thanks u @gb2017
– jasmine
Nov 20 '18 at 21:30
add a comment |
thanks u @gb2017
– jasmine
Nov 20 '18 at 21:30
thanks u @gb2017
– jasmine
Nov 20 '18 at 21:30
thanks u @gb2017
– jasmine
Nov 20 '18 at 21:30
add a comment |
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