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$A_1=\{ \text {closed unit disk in plane}\}$ $A_2=\{(1,y):y\in \mathbb{R}\}$ $A_3=\{(0,2)\}$ 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?

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    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.)2012-06-13
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    which theory or result we are using here could you please tell me?2012-06-13
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    Too see this, you only need the definition of function. (I am assuming that by closed unit disc you mean $A_1=\{(x,y)\in\mathbb R^2; x^2+y^2\le 1\}$.)2012-06-13
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    ya thats true....2012-06-13
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    @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.2012-06-13

1 Answers 1

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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}$

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    No, 2) is ok and you to apply Urysohn2012-06-13
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    Sorry, I don't know what you mean, can you please clarify?2012-06-13
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    1. If $a_1=a_2=a_3$ then we may choose $f$ as a constant function.2012-06-13
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    I like the updated answer much more. Note also that "Urysohn for metric spaces" is much easier than the full lemma.2012-06-13
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    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 sharp2012-06-13
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    Now I understand your confusion, the $a_i$ are the values of $f$ and not points in the plane2012-06-13
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    @SimonMarkett Ok! Now I see my mistake! I somehow read $f(x)=a_j$ where $a_j\in A_j$....! Sorry :)2012-06-13
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    I am sorry for taking your time answering my stupid comments.2012-06-13
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    Now I have rethought it all. 4 is true as you say (but still you do not need Urysohn :) But as I see it 2 is still true, because the function may be constantly equal to $a_1$...2012-06-13
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    No worries. It can happen to all of us. Whether you need or don't need Urysohn somehow depends on how you see the question. You can easily define other sets $A$ such that writing down a function explicitely can get really tedious. And urysohn tells you that it essentially still works in the same way. But yeah, here it's probs way to much.2012-06-13
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    Actually I didn't say that 2) was wrong. I did say 2) is stronger than 4) and 4) suffices.2012-06-13