metric space

penelope

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Show that for any metric space X , the set X \ {x} is open in X .i know that an open set in X is an open ball but i don't really get it
 
Show that for any metric space X , the set X \ {x} is open in X .i know that an open set in X is an open ball but i don't really get it

since {x} is closed set (becoz closure of {x} ={x})
therefore its complement that is X\{x} is open set......
 
Show that for any metric space X , the set X \ {x} is open in X .i know that an open set in X is an open ball but i don't really get it


If yX{x}\displaystyle y\in X\setminus\{x\} then yx\displaystyle y\ne x so d(x,y)>0 .\displaystyle d(x,y)>0~.

(yX{x})\displaystyle (\forall y\in X\setminus\{x\}) define δy=d(x,y)2\displaystyle \delta_y=\frac{d(x,y)}{2}

X{x}=yX\{x}B(y;δy)\displaystyle X\setminus\{x\}=\bigcup\limits_{y \in X\backslash \left\{ x \right\}} {B(y;\delta _y )} .

The union of open sets is open.
 
Show that for any metric space X , the set X \ {x} is open in X .i know that an open set in X is an open ball but i don't really get it
Unfortunately, what you "know" isn't true. In a metric space, every open ball is an open set, but the other way is not true- not every open set is an open ball.

Let p be in X\ {x}. Let d=δ(p,x)\displaystyle d= \delta (p, x). Show that the ball, centered on p, with radius δ/2\displaystyle \delta/2 is in X\{x}.
 
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