Continuous function takes discrete value on disjoint closed sets

Asked Oct 21 '20 at 15:04

Active Oct 21 '20 at 15:04

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Let $(X,d)$ be a metric space. $A,B$ be disjoint closed subsets of $X$. Then there must exists a continuous function $f$ on $X$, with $f(A)=0$, $f(B)=1$.

My idea is to take combination of distance function to $A, i.e$ $1-d(x,A)$. Then the function is continuous and take value 0 on $A$. But I got stuck how to take care of the values in $B$.

asked Oct 21 '20 at 15:04

xyz

1,457

Urysohn's lemma surely? – copper.hat Oct 21 '20 at 16:07
1

How about ${d_A \over d_A+d_B}$? – copper.hat Oct 21 '20 at 16:49
https://math.stackexchange.com/questions/153765/continuous-function-on-metric-space?rq=1 – copper.hat Oct 22 '20 at 20:27

Continuous function takes discrete value on disjoint closed sets

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