Exercise 1.5

For any set $$X$$, we can define $$d$$ by $$d(x,x)=0$$ and $$d(x,y)=1$$ if $$x \neq y$$. Check that $$d$$ is a metric. This metric is called the discrete metric on $$X$$.

Tony bruguier
Items 1 and 2 are obvious by definition. What remains to be proven is item 3.

If $$x=z$$, then $$d(x,z)=0$$ and nothing is to be proven, as the left-hand side will always be positive.

If $$x \ne z$$, then $$d(x,z)=1$$. We have three cases, either $$y = x$$ and $$y \ne z$$, or $$y \ne x$$ and $$y = z$$, or $$y \ne x$$ and $$y \ne z$$. But for these three cases, we have $$d(x,y) + d(y,z)$$ equal to 1, 1, or 2; thus proving the result.

Tony Bruguier 23:59, February 25, 2011 (UTC)

'''Great! I see you did this before we started the course - now that we're underway, we're asking that people post to the current week's thread in /r/topology2011 whenever they attempt a solution to drive discussion. Thanks! --Steven.clontz 03:15, March 2, 2011 (UTC)'''