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I'm trying to prove the continuity of the translation map: $f:\mathbb R \to \mathbb R$, defined by $f(x)=x+k$, where $k$ is an integer.

I know that the preimage of an open interval is an open interval. Since an open set of $\mathbb R$ is defined as an union of open intervals, I think the question becomes a little more complex.

Can you help me to prove the continuity of this map?


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up vote 2 down vote accepted

Since the inverse image under $f$ of an open interval is an open interval, and inverse images preserve unions, the inverse image under $f$ of an open set is open. When I say that inverse images preserve unions, I mean that $$f^{-1}\left[\bigcup\mathscr{U}\right]=\bigcup_{U\in\mathscr{U}}f^{-1}[U]\;.$$ Thus, if each $U\in\mathscr{U}$ is an open interval, then each $f^{-1}[U]$ is an open interval, and $f^{-1}\left[\bigcup\mathscr{U}\right]$ is a union of open intervals and therefore an open set.

You can also very easily use the $\epsilon$-$\delta$ definition of continuity: for each $\epsilon>0$ just take $\delta=\epsilon$, and observe that if $|x-x_0|<\epsilon$, then $|f(x)-f(x_0)|<\epsilon$. (This actually shows that $f$ is not just continuous, but uniformly continuous.)

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In order to prove the homeomorphism, it suffices to use the same argument to prove the continuity of inverse function $f(x)=x-k$? thank you for your answer – user42912 Dec 26 '12 at 13:29
It may be worth mentioning that by this argument a map is continuous if the preimage of every basis-element is open. – Dave Dec 26 '12 at 13:29
@Rafael: You’re welcome. You don’t even need to make a separate argument: if we let $f_k$ be the function that sends $x$ to $x+k$, then $f_k^{-1}=f_{-k}$ is just another function of the same type, so you’ve already proved that it’s continuous. – Brian M. Scott Dec 26 '12 at 13:32

Suppose $U\subseteq \mathbb{R}$ is open. Then $S=f^{-1}(U)=\left\{x\in \mathbb{R}:x+k\in U\right\}$. We must prove $S$ is open

If $x\in S$ then $x+k\in U$ and so by the open-ness of $U$, $$\exists r>0: (x+k-r,x+k+r)\subseteq U$$ Thus $$y\in(x-r,x+r)\implies y+k\in U\implies y\in S$$ Therefore, $(x-r,x+r)\subseteq S$ and so $S$ is open.

It remains to prove $$y\in (x-r,x+r)\implies y+k\in(x+k-r,x+k+r)$$ but that is simple

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