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Let $I_n = (-n,n)×\cdots×(-n,n)$ in $\mathbb{R}^k$. ($n$ is a positive integer)

Fix $x\in I_n$ Let $L=\{\min\{n-x(i),x(i)+n\} \mid i\in k\}$. Since $L$ is finite and is a subset of $\mathbb{R}$, it is well-ordered. Let $z$ be the least element of $L$.

Let $y\in N_z(x)$. (neighborhood of x with radius z)

I'm trying to show that for every $i\in k, -n to show that $x$ is an interior point of $I_n$, but it didn't work well.. Help

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    @Mercy natural number $1≦i≦k$ when $k$ is a natural number2012-07-29

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You have $n-x_i \geq z >0$ and $n+x_i \geq z >0$ for all $i$. Furthermore, you have $|y_i-x_i| < z$, which is equivalent to $-z < y_i-x_i < z$.

Then you have $n-y_i = n-x_i +y_i-x_i \geq z + y_i-x_i > z -z= 0$. Similarly, you have $n+y_i = n+x_i - (y_i-x_i) \geq z - (y_i-x_i) > z -z = 0$. Hence $y_i \in (-n,n)$ for all $i$. It follows that $y \in I_n$.

NOTE:

I need to elaborate on a point above. In the statement $y\in N_z(x)$, I was taking the neighborhood to be $\{y \ | \ |y_i-x_i| < z \in \forall i \}$. This is a perfectly valid neighborhood (using the infinity norm $\|\cdot\|_{\infty}$), but may not agree with your norm. A more usual norm would be the 2-norm $\|x\|_2 = \sqrt{\sum_i x_i^2}$, in which case you would need to adjust the neighborhood slightly, since the relevant bound is $\|x\|_2 \leq \sqrt{n} \|x\|_{\infty}$. If you are using the 2-norm, then you should choose $y \in N_{\frac{z}{\sqrt{n}}} (x)$, in which case it will then be true that $-z < y_i-x_i < z$.

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    @Katlus: Actually, that's a good point and one I glossed over, I will add a note to the solution.2012-07-29