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After reading the definition of a continuous map on general topological spaces, my question is the following: Suppose $f$ is continuous from $\mathbb R$ to $\mathbb C$ given by $x \mapsto e^{ix}$. Now this makes since the circle $S_1$ is closed in the ambient space $\mathbb C$ and so is $\mathbb R$. But $f$ is supposed to be continuous on any subset of $\mathbb R$ as well. Now if I restrict $f$ to $(-4\pi, 4\pi)\to \mathbb C$ given by the same mapping, I get the same closed circle, but its preimage is now not closed. Or am I supposed to be viewing this interval with a subspace topology and considering it as the whole space, in which case the interval is both open and closed. Finally, if this is the case, am I correct in stating that a continuous function may map an open set to a closed one if the mentioned open set is also closed?

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    $\begingroup$ notice the maths formatting use '$' symbols. $\endgroup$ – Ittay Weiss Sep 28 '13 at 9:06
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You need to distinguish between the two different functions $f:\mathbb R \to \mathbb C$ given by $f(x)=e^{ix}$ and its restriction $g=f|_{(-4\pi, 4\pi)}$. The definition of continuity is sensitive to the domain and its topology. So, both of these functions are continuous since the inverse image of an open in the codomain $\mathbb C$ is open in the topology of the domain. The domain for $f$ is $\mathbb R$, but the domain for $g$ is $(-4\pi,4\pi)$, and indeed $(-4\pi, 4\pi)$ is closed in this space.

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  • $\begingroup$ thanks, that makes perfect sense, but then why necessarily restrict to an open interval? How about taking a subset that is closed in $R$ but throw it into the subspace topology set and call it open. I guess i am trying to say is the interval required to be open in the ambient space? $\endgroup$ – masszz Sep 28 '13 at 9:17
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    $\begingroup$ @sergey It's not required to be open. The restriction of a continuous function to any subspace is continuous. $\endgroup$ – Daniel Fischer Sep 28 '13 at 9:20
  • $\begingroup$ So to prove the continuity in the restriction: Let $X$ be the original space, $U$ be the subspace and $f(U) = V.$ Let $W \in V$ be open, then $f^{c}(W)$ is open in $X$ since $f$ is cont. on it. *But now how does this imply the openness of $f^{c}(W)$ in $U$ if we don't know apriori the topology set on $U.$? * $\endgroup$ – masszz Sep 28 '13 at 9:47
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Here I respond only to your final sentence:

Question: Can a continuous function map a set that is closed and open to a closed one?

Answer: Sure. Consider $f: \mathbb{R} \rightarrow \mathbb{R}$ defined by $x \mapsto 1$.

Then the closed and open set $\mathbb{R}$ is mapped to the closed set $\{1\}$.

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Indeed, restricting the domain formally gives you a function $\tilde f: (-4\pi,4\pi) \to \Bbb C$. Then indeed $\tilde f$ will be continuous.

As to your other point, we have no control on what a continuous mapping maps things to. We only know that the preimage of any open is again open.

But even in the present case, a part of the image $S^1$ in $\Bbb C$ (say the image of an open set) will neither be open nor closed in $\Bbb C$, because it does not contain all its limit points (hence not closed), but has empty interior (hence cannot be open).

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  • $\begingroup$ Thinking about continuous mappings in terms of images, rather than preimages, can still lead to somewhat interesting considerations. A nice example is: Characterizing Continuity. Daniel J. Velleman. The American Mathematical Monthly, Vol. 104, No. 4 (Apr., 1997), pp. 318-322, jstor.org/stable/2974580. $\endgroup$ – Benjamin Dickman Sep 28 '13 at 9:05
  • $\begingroup$ but we can state that a non-open set can not be mapped to an open set. Correct? $\endgroup$ – masszz Sep 28 '13 at 9:19
  • $\begingroup$ @sergey That's not correct. For example with a constant mapping to the discrete topology, any set has an open image. The preimage of any open set is either empty or the whole space, so the constant mapping is nonetheless continuous. $\endgroup$ – Lord_Farin Sep 28 '13 at 9:26
  • $\begingroup$ @Lord_Farin ok, just to clarify, when we say "restriction of f to the set U", we do not mean it in the sense that you just did. We mean U becomes the whole space for f. $\endgroup$ – masszz Sep 28 '13 at 9:38
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    $\begingroup$ @BenjaminDickman That's nice, thanks for the link. $\endgroup$ – Lord_Farin Sep 28 '13 at 9:50

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