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Given a set of functions $M=\lbrace f_k \rbrace$, where $f_k = \sin (kx)$, prove that M is closed and bounded.

I have been given a norm: $$\| f\| = \left[ \int_{0}^{2 \pi} f (x)^2 dx \right]^{1/2}$$ (on the space of continuous, real functions from $[0,2\pi]$ to $\mathbb{R}$), but no metric or topology. I don't see how I can do this.

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    What norm had you been given? Also, a norm automatically gives you a metric ($\therefore$ also a topology). – 2017-01-09
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    The norm was on the space of continuous, real functions from $[0, 2 \pi] \rightarrow \mathbb{R}$, defined as: $|| \cdot || = \left[ \int_{0}^{2 \pi} f (x)^2 dx \right]^{1/2} $ – 2017-01-09
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    use the open ball topology, defining $B(r, \epsilon) := \{x \in X : ||x-r|| < \epsilon\}$ – 2017-01-09
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    Thanks, man! I did it now. – 2017-01-10

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General remarks

  • A norm induces a metric: $d(x,y) = \|x-y\|$.
  • A metric induces a topology, via neighborhoods $N(x,r) = \{y:d(x,y)

Specific set

Boundedness amounts to a direct computation: $\|f_k\| = \pi^{1/2}$

The reason the set is closed is that its points are uniformly far apart; as a result, the set has no limit points. To justify this, compute $\|f_k-f_j\| = (2\pi)^{1/2}$ for $k\ne j$.

The computations are based on double-angle and product-to-sum identities.