Show A neither open nor closed in (R^2) , where A={(x,o): -1<x<1}
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Originally Posted by rqeeb Show A neither open nor closed in (R^2) , where A={(x,o): -1<x<1} Sorry to tell you that question is total nonsense. If $\displaystyle x=0.5$ the set $\displaystyle (x,0)$ does not exist. Please reread and repost a correct question.
Originally Posted by Plato ... Please reread and repost .... Indeed.
Originally Posted by rqeeb Show A neither open nor closed in (R^2) , where A={(x,o): -1<x<1} Just use the definitions. Not open means there exists a point in R^2 such that the ball around this point (for any E > 0) contains a point not in A.
Originally Posted by TheChaz Indeed. but he Excersize in the book is written by this way:in $\displaystyle {R}^2}$ show that: $\displaystyle A=\{(x,0):\{-}\1<x<1\}$ is neither open nor closed.
Originally Posted by rqeeb but he Excersize in the book is written by this way:in $\displaystyle {R}^2}$ show that: $\displaystyle A=\{(x,0):\{-}\1<x<1\}$ is neither open nor closed. My first reply was a cheeky way of informing P that HE misread the question. My second reply is the direction you should go.
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