Is the solution to Q2 incorrect? The solution give the topology, T = {phi, X, {a}, {b}, {a, b}, {b, c}, {a, d, e}} Here, {a, b} U {b, c} = {a, b, c} does not belong to T. Hence, T can't be a topology.
Sir, as here ot was said that finite intersection, then why not you consider empty intersection? If someone consider empty intersection, Will be there any problem?
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At 5:48 (a,b,c)&(a,b,d,e) also will be in Topology T.
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At 5:47
T also has a member {a,b,c}
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Grazie per il video. Il concetto chiamato in causa è proprio quello di TOPOLOGIA INDOTTA.
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Sir finite intersection means
Is the solution to Q2 incorrect?
The solution give the topology, T = {phi, X, {a}, {b}, {a, b}, {b, c}, {a, d, e}}
Here, {a, b} U {b, c} = {a, b, c} does not belong to T. Hence, T can't be a topology.
Also {a,b,d,e}
We have to form the topology, so T must have (a, b, c ) and (a, b, d, e)
Sir in question ,union of {a},{c} is {a,c} which is not present in T ?? So it is not a sub-base ??
Question is wrong,{a,c} must present in T
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IN T NEED ONE SET (a,b,c) to be a Topology
Sir, as here ot was said that finite intersection, then why not you consider empty intersection? If someone consider empty intersection, Will be there any problem?
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Sir, in question number 2, there is not {a,b,c} in topology. Without this set T can't be a topology.
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