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life is maths
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Hi, my instructor left this as an exercise, but I got confused in the second part. Could you please help me?
cl(A[itex]\cap[/itex]B)[itex]\subseteq[/itex]cl A [itex]\cap[/itex] cl B
But the reverse is not true. Prove this and give a counterexample on the reverse statement.
My attempt:
If x[itex]\in[/itex] A[itex]\cap[/itex]B, then x[itex]\in[/itex] cl(A[itex]\cap[/itex]B)
x[itex]\in[/itex] A and x[itex]\in[/itex] B [itex]\Rightarrow[/itex] x[itex]\in[/itex] cl(A) and x[itex]\in[/itex]cl(B). Hence,
cl(A[itex]\cap[/itex]B)[itex]\subseteq[/itex]cl A [itex]\cap[/itex] cl B
Does this proof have any flaws? It is an easy one, I guess, but I feel a bit confused. And I do not understand why the reverse is wrong. Can't we use the same method? Thanks for any help.
cl(A[itex]\cap[/itex]B)[itex]\subseteq[/itex]cl A [itex]\cap[/itex] cl B
But the reverse is not true. Prove this and give a counterexample on the reverse statement.
My attempt:
If x[itex]\in[/itex] A[itex]\cap[/itex]B, then x[itex]\in[/itex] cl(A[itex]\cap[/itex]B)
x[itex]\in[/itex] A and x[itex]\in[/itex] B [itex]\Rightarrow[/itex] x[itex]\in[/itex] cl(A) and x[itex]\in[/itex]cl(B). Hence,
cl(A[itex]\cap[/itex]B)[itex]\subseteq[/itex]cl A [itex]\cap[/itex] cl B
Does this proof have any flaws? It is an easy one, I guess, but I feel a bit confused. And I do not understand why the reverse is wrong. Can't we use the same method? Thanks for any help.