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Are there two topological space $T_1$ and $T_2$ in which a continuous constant function sends an open set of the first topological space to a non-open set of the second?

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Every nonempty $T_1$, and every $T_2$ which does not have the discrete topology, have this property. –  Nate Eldredge Nov 25 '11 at 15:21
    
assuming you mean "non-constant", so I fixed it. –  GEdgar Nov 25 '11 at 16:08
    
@GEdgar: No, I know examples of non-constant functions that do not preserve openness. What I really wanted to know was the existence of a constant function that sends a "pure" open set (one that is not closed at the same time) in a non-open one. As you can see, it seems to be impossible because the preimage of a closed set must be closed. I believe the only cases are those in which the set in $T_1$ is clopen, as kahen's example. –  user14174 Nov 25 '11 at 18:19
    
OK back to "constant". –  GEdgar Nov 25 '11 at 20:01
    
You should try to avoid using $T_1$ and such for spaces in topology. These are used quite often to describe properties of the space. –  Asaf Karagila Nov 25 '11 at 21:30

2 Answers 2

up vote 2 down vote accepted

Yes. $\mathbb R \owns x \mapsto c \in \mathbb R$ for any real number $c$.

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yes, it's the first function that came in my mind, thanks for confirming. But now I see I have posted a bad question, in the sense that my real interest was in open and just open sets, that is: (open, not close) -> (not open). –  user14174 Nov 25 '11 at 15:24
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@Lmn6: I don't understand what you mean. This map sends every nonempty open set to the set $\{c\}$, which is not open. Isn't that what you wanted? –  Nate Eldredge Nov 25 '11 at 20:33
    
@NateEldredge: Nevermind, it was a bad reasoning that I was applying. I try to explain. Since $f:x \in \mathbb{R} \mapsto c$ is continuous, then $f^{-1}(\{c\})$ is closed. Now $\mathbb{R}$ is closed and that's ok. But I did not immediately understand at first glance why, for example, when picking an open interval $A=(a,b)$ and considering the restriction to that interval, I got $f^{-1}(\{c\})=A$ open... when it must have been closed by continuity hypothesis. In considering restrictions one has to change the topological space, so A would be clopen and continuity holds. –  user14174 Nov 25 '11 at 21:11

For a trivial example let $X$ by $\mathbb{N}$ with the initial segment topology, and let $Y$ be $\mathbb{N}$ with the final segment topology. That is, $U\subseteq X$ is open iff $m\le n\in U$ implies that $m\in U$, and $U\subseteq Y$ is open iff $m\ge n\in U$ implies that $m\in U$. Let $f:X\to Y:n\mapsto n$. For any $A\subseteq X$, $$f[A]\text{ is }\begin{cases} \text{closed,}&\text{if }A\text{ is open in }X\;;\\ \text{open,}&\text{if }A\text{ is closed in }X\;;\text{ and}\\ \text{neither open nor closed,}&\text{if }A\text{ is neither open nor closed in }X\;. \end{cases}$$

Of course $X$ and $Y$ are only $T_0$, so they’re not especially nice spaces. However, we can’t actually do much better if $f$ is to be a surjection. Suppose that $f:X\to Y$ is surjective, where $X$ and $Y$ are both $T_1$. Then for each $x\in X$, $X\setminus\{x\}$ is open in $X$, but $f[X\setminus\{x\}]$ is either $Y\setminus\{f(x)\}$ or $Y$, both of which are open in $Y$.

If $f$ needn’t be surjective, there are easy examples with nice spaces. For instance, let $\mathbb{P}$ be the space of irrationals with the usual topology, and let $f:\mathbb{P}\to\mathbb{R}:x\mapsto x$. Then the only open set sent to an open set by $f$ is the empty set.

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