How to find accumulation points and open sets in a sequence?

[ShengyaoLiang]

1) give an example of a sequence (An) of open sets such that the intersetion

∩ An is not open.
n in N


2) If A is a nonempty set of real numbers bounded below with no minumum, then infA is an accumulation point of A.

Could somebody gives some hint on those questions? Thank you very much...i have no i dear on the 2nd prove...

 

[d_leet]

2) If A is a nonempty set of real numbers bounded below with no minumum, then infA is an accumulation point of A.

What is your definition of an accumulation point? The definition I am familiar with is the one in Royden's real analysis text that says a point x is an accumulation point of the set E if it is a point of closure of the set E-{x}. In your case however since A has no minimum then infA cannot belong to A (suppose it does belong to A, and A has no minimum, then there exists a element m in A such that m is smaller than infA, but this contradicts that infA is a lower bound of A, hence infA does not belong to the set A.) then obviously A=A-{infA} so you would only need to show that infA is a closure point of A, or that every neighbourhood about infA contains a point in A.

 

[SiddharthM]

intersect decreasing neighborhood whose diameter goes to zero, make sure all neighborhoods have a common point you then get their intersection to be a single point which is not open unless you are using the discrete metric/topology.

 

[ShengyaoLiang]

Thank You Very Much~~~~hoho