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True or False:

There exists a continuous function $f : \mathbb{R}^2 → \mathbb{R} > $such that $f ≡ 1$ on the set $\{(x, y) \in \mathbb{R}^2 : x ^2+y^2 =3/2 \}$ and $f ≡ 0$ on the set $B∪\{(x, y) \in \mathbb{R}^2: x^2+y^2 ≥ 2\}$ where B is closed unit disk.

I think this is just a stratightforward application of Urysohn's Lemma as metric spaces are normal so by Urysohn's Lemma says that disjoint closed subsets can be separated by continuous function.

I hope I am not missing something. Topology can be weird sometimes!!!

Thanks in advance.

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    $\begingroup$ That's true you can use Urysohn's Lemma. You just need to show that these sets are closed and disjoint. $\endgroup$ – Yanko Jan 11 at 20:26
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    $\begingroup$ You can use Urysohn's Lemma, yes. It seems kind of like using a nuke to kill ants, though, especially when it's pretty clear how to just write down what $f$ is explicitly. $\endgroup$ – user3482749 Jan 11 at 20:37
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I think this is just a straightforward application of Urysohn's Lemma as metric spaces are normal so by Urysohn's Lemma says that disjoint closed subsets can be separated by continuous function.

Right. As you say and as Yanko agrees, you can use Urysohn's Lemma; then it just remains to show the sets are closed and disjoint.

In case you don't like Ursyohn's Lemma -- or just for fun -- we can define the continuous function ourselves. It doesn't turn out to be so hard in this particular case. Define $f: \mathbb{R}^2 \to \mathbb{R}$ by $$ f(x,y) = 2(2 - x^2 - y^2). $$

Now, this is a polynomial, so it's continuous. And on the set $A$, $f(x,y) = 2(2 - \tfrac32) = 2 \cdot \tfrac12 = 1$. And on the set $B$, $f(x,y) = 2(2 - 2) = 0$.

If $f$ has to be positive, you can instead make $f(x,y) = \max(0, 2(2-x^2 - y^2))$.

Topology can be weird sometimes!!!

I agree :)

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