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Prove that the open ball $B(\mathbf{p}, \delta)=\{\mathbf{x} :\|\mathbf{x}-\mathbf{p}\| < \delta\}$ is aconvex set. I Hint: Use the Triangle Inequality.

It follows that $z \in B(p, \delta)$ and $B(p, \delta)$ is convex.

Calculus 3

Chapter 8

The Geometry of Vector Spaces

Section 4

Hyperplanes

Vectors

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Lectures

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In mathematics, a vector (…

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If $c \in \mathbb{R}$ and …

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00:54

Prove the triangle inequal…

01:25

Use (6) to prove the Trian…

03:13

The Triangle Inequality fo…

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the successes suggest stirs in hand. Now use the triangular inequality so tranquil. Earning a quality says that the absolute value off a plus B is in less than or equal than absolute body. Oh, very plus the absolute value be for every A m B. So the sheen says. Try that for why and X minus y. So if a CE Why NBC's X minus y we have a thing for the sum this expression less than a recall than absolute value of y plus X minus y absolute back. Now why Blast X minus y is X. So now if we subtract both sides, Nick and the absolute value why we get ex miners absolute value, why is less than or equal than X minus y? That's when we wanted to croak.

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The number 2 is also the smallest & first prime number (since every other even number is divisible by two).

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