Thread: help with proving a set to be compact

let K be compact in R^n. show that a + K is also compact, where a + K := { a + x : x in K}


what I was thinking is that since K is compact there exists a finite open cover of K. a is a single element and so can also be covered
by a finite open cover and the union of the two covers is also finite so a + K is contained within a finite open cover and hence compact.
am I close?

I don't see that approach working.

You could try doing it this way:

$\displaystyle \text{For each } a \in \mathbb{R}^n, \text{ define }f_a : \mathbb{R}^n \rightarrow \mathbb{R}^n \text{ by } f_a(x) = x + a.$

Can you show that f is continuous? If so, that's actually all you need, but you could then quickly prove much more:

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Maybe this is enough to guide you, maybe it isn't. But think about:

$\displaystyle \text{What are } f_a \circ f_{(-a)} \text{ and } f_{(-a)} \circ f_{a}?$

$\displaystyle \text{So, for each } a \in \mathbb{R}^n, \text{ have that } f_a \text{ is ???, with ??? = } f_{(-a)}.$

$\displaystyle \text{But } f_{(-a)} \text{ is also c???, and thus } f_a \text{ is a h???.}$

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You could also try doing it this way, which amounts to about the same thing while requiring more work:

$\displaystyle \text{Prove: For all } a \in \mathbb{R}^n, \text{ and } U \subset \mathbb{R}^n, \ U \text{ is open iff } a + U \text{ is open.}$