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Let $S = \{ (a_1, a_2, a_3)\in \mathbb{R}^3 | \ a_1 + a_3^2\cdot\sin( a_1 +a_2) \geq a_3 \}$

then, how can I, show that S is closed under Euclidean Metric.

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@ram: Should that be $\sin(a_1+a_2)$ or $\sin(2a_1)$? –  Dennis Gulko Oct 18 '12 at 10:49
    
yes !!! I copied incorrectly, Now I correct it.. Thanks –  ram Oct 18 '12 at 10:50
    
Can you say a little bit about what you know? It would help to know if this is for an intro to analysis class, a point set topology class, or some other class. Also, can you say a little on what you've tried? Both will help us help you better. –  Aaron Oct 18 '12 at 11:07

3 Answers 3

up vote 6 down vote accepted

A function $f\colon \mathbb{R}^3 \to \mathbb{R}$ is continuous

iff

the inverse image of every open set $U\subset \mathbb{R}$ is open in $\mathbb{R}^3$

iff

the inverse image of every closed set $A\subset \mathbb{R}$ is closed in $\mathbb{R}^3$.

Now relabeling $a_1,a_2$ and $a_3$ as $x,y,z$, moving $a_3$ to the LHS, we get that $S=\{(x,y,z) \in \mathbb{R}^3 \, : \, f(x,y,z) \ge 0 \},$ where $f(x,y,z)= x + z^2 \sin(x+y) - z$.

Finally, $f$ is continuous, and $S$ is the inverse image, under $f$, of the closed set $[0,\infty)$, hence $S$ is closed.

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Hint: I would suggest to rename the coordinates to say $a,b,c$, and then consider convergent sequences $(a_n,b_n,c_n)\in S$, and prove that the limit still satisfies the condition of $S$, by arguing with continuity ($\implies$ sequential continuity) of multiplication, squaring and sine (too easy, but wouldn't hold with $>$ instead of $\ge$).

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Here's a hint for an approach slightly different from Berci.

Depending on what definition of continuous you're using, it is straight forward that the inverse image of a closed set is closed. Show that you can rewrite your set as $f^{-1}([0,\infty))$ for some continuous function $f$.

Note, to show that $f$ is continuous, you will want to use that compositions of continuous functions are continuous, so you are reduced to showing that a few basic functions $(+, \cdot, \sin)$ are continuous.

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