# does there exist a continuous function

$A_1=\{ \text {closed unit disk in plane}\}$ $A_2=\{(1,y):y\in \mathbb{R}\}$ $A_3=\{(0,2)\}$ We need to confirm: there exist always a continuous real valued function $f$ on $\mathbb{R}^2$ such that $f(x)=a_j$ for $x\in A_j$ $j=1,2,3$

$1$. Iff atleast two of these number are equal.

$2$. all are equal.

$3$. for all values of these 3 numbers.

$4$.iff $a_1=a_2$

Is some how I need to use Urysohn's lemma here?

-
Since $A_1\cap A_2\ne\emptyset$, you need $a_1=a_2$. (Even if you don't require the function to be continuous.) – Martin Sleziak Jun 13 '12 at 11:35
which theory or result we are using here could you please tell me? – La Belle Noiseuse Jun 13 '12 at 11:36
Too see this, you only need the definition of function. (I am assuming that by closed unit disc you mean $A_1=\{(x,y)\in\mathbb R^2; x^2+y^2\le 1\}$.) – Martin Sleziak Jun 13 '12 at 11:37
ya thats true.... – La Belle Noiseuse Jun 13 '12 at 11:59
@Mex No, you do not need Urysohn's lemma. In each case try to construct such a function or prove that no such function can exist. – AD. Jun 13 '12 at 12:35

As discussed in the comments you need to have $a_1=a_2$ (note that $(1,0)\in A_1\cap A_2$), so the only options are 2) and 4), where 2) is the stronger assumption. We can show however that 4) suffices.

Indeed we may apply Urysohn to the sets $A=A_1\cup A_2$ and $B=A_3$. Both sets are closed and disjoint, moreover $\mathbb R^2$ is normal. If $a_1=a_3$ choose $f\equiv a_1$. So assume $a_1\neq a_3$.

By Urysohn there exists a continuous function $f:\mathbb R^2\to [0,1]$ with $f(A)=0$ and $f(B)=1$. Postcompose this map with the canonical homeomorphism $[0,1]\to [a_1,a_3]$ if $a_1< a_3$ or the strictly decreasing homeomorphism $[0,1]\to [a_3,a_1]$ if $a_3< a_1$. We are done.

Edit: Maybe this is actually a bit of an overkill. Since your sets are given explicitely you can just define $f(x,y)=\begin{cases}a_1& \text{ if$x\leq 1$}\\ a_3&\text{ if$x\geq 2$}\\ a_1+(x-1)(a_3-a_1)&\text{ if$1\leq x\leq 2$}\end{cases}$

-
No, 2) is ok and you to apply Urysohn – AD. Jun 13 '12 at 12:38
Sorry, I don't know what you mean, can you please clarify? – Simon Markett Jun 13 '12 at 12:39
1. If $a_1=a_2=a_3$ then we may choose $f$ as a constant function. – AD. Jun 13 '12 at 12:41
I like the updated answer much more. Note also that "Urysohn for metric spaces" is much easier than the full lemma. – Dylan Moreland Jun 13 '12 at 12:46
No, $a_1$ and $a_2$ have to be equal, otherwise it doesn't work for points in the (non-trivial) intersection $A_1\cap A_2$. Therefore 1) and 3) are to weak. 4) Is true, but stronger than 2). 2) is sharp – Simon Markett Jun 13 '12 at 12:47