Topology - continuous functions

Let $X = \{1, 2, 3, 4, 5\}$ with topology $\{\emptyset, X, \{1\}, \{3, 4\}, \{1, 3, 4\}\}$, and let $Y = \{A, B\}$ with topology $\{\emptyset, Y, \{A\}\}$. Find all continuous functions from $X$ to $Y$. How many functions from $X$ to $Y$ are there altogether?

This is what I have thus far:

Let $(X,\tau)$ and $(Y,\tau^\prime)$ be topological spaces. A function from $X$ to $Y$ is said to be continuous if given any open subset $U$ of $Y$, then $f^{-1}(U)$ is an open subset of $X$.

a. Continuous functions from $X$ to $Y$ can be characterized as those that map open subsets of $X$ to open subsets to $Y$. In total, there exist $8$ continuous functions from $X$ to $Y$: \begin{align*} f&: \{{1}\} \rightarrow A\\ f&: \{{3, 4}\} \rightarrow A \\ f&: \{{1, 3, 4}\} \rightarrow A \\ f&: X \rightarrow A \\ f&: \{{1}\} \rightarrow Y \\ f&: \{{3, 4}\} \rightarrow Y \\ f&: \{{1, 3, 4}\} \rightarrow Y \\ f&: X \rightarrow Y \end{align*}

My question is: how do I handle the empty set? I know that if I am mapping to or from the empty set, $X \times Y = \emptyset$, i.e. $\emptyset$ is the only subset of $X \times Y$. So, we shouldn't be able to map $\emptyset$ to a nonempty set, correct? And, should we be able to map a nonempty set to $\emptyset$? Thanks for the help y'all!

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These $f^i$ are not even functions –  Stefan Hamcke Feb 10 '14 at 16:56
No, we shouldn't be able to "map $\emptyset$ to a nonempty set", neither should we be able to "map a nonempty set to $\emptyset$." But I think the way you're looking at this is slightly off: in the first instance all we want are functions $f$ from $X$ to $Y$, i.e. all you need to do is say what $1,2,3,4,5 \in X$ each map to. Then all you need to do is check if the set of things in $X$ that map to $A \in Y$ (note: $A\in Y$, not $\{A\} \subset Y$) is open. In your list of continuous functions $f^i$, you're doing something funky with the sets, and that's the off bit. –  Josh Chen Feb 10 '14 at 17:02
Also, what do you mean by $X\times Y = \emptyset$? –  Josh Chen Feb 10 '14 at 17:04
I believe that describes what I did--does it not? Yes, I know that I labeled my functions strangely. I wanted to number them, but should not have done it that way in retrospect. –  kathystehl Feb 10 '14 at 17:21
I mean that the function X to Y is a subset of X x Y, satisfying certain conditions. Is this not how one usually says that? –  kathystehl Feb 10 '14 at 17:22

The fact that the preimage of an open is open IS NOT equivalent to the fact that $f$ maps open sets to open sets. Such maps are called open maps. There exist continuous maps that are not open, open maps that are not continuous, open continuous maps and maps which are neither continuous nor open!

To find continuous maps from $X$ to $Y$ in your case amounts exactly to find all the possible preimages of $A$. Since $Y$ has only two elements, the preimage of $A$ determines uniquely the map. So there are exactly 5 continuous maps $X\to Y$. Two of them are the constant ones, the third is the one mapping $1$ to $A$ (and everything else to $B$), the fourth is the one mapping $3$ and $4$ to $A$ and the fifth maps $1$,$3$ and $4$ to $A$.

You don't have to care about $\varnothing$, because its preimage under any map is again $\varnothing$ which has to be open by definition of topology. In the same way, you don't have to care about the preimage of $Y$, which is always $X$.

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What are the two "constant ones?" If I am disregarding any mapping of the empty set or of X to Y, what is left over that you are denoting as the two constant ones? Thanks for the clarification. –  kathystehl Feb 10 '14 at 17:02
One is the map sending everything to $A$, the other one is the map sending everything to $B$. –  Ferra Feb 10 '14 at 17:03
But mapping everything to B would not be a constant function since B is not an open set, correct? –  kathystehl Feb 10 '14 at 17:09
constant map=mapping everything to a point. Mapping everything to $B$ is continuous, since the preimage of $A$ is the empty set which is open. –  Ferra Feb 10 '14 at 17:11

Edit: First, let me clarify some notation and definitions that I will be using here. Suppose we have a set $X$ and a set $Y.$ Then we say that a subset $f$ of $X\times Y$ is a function from $X$ into $Y$--denoted $f:X\to Y$--if and only if for each $x\in X$ there is a unique $y\in Y$ such that $\langle x,y\rangle\in f$--and we call this unique $y$ by the name $f(x).$ Given $S\subseteq Y,$ the preimage of $S$ under $f$ will be denoted and defined by $$f^{-1}[S]:=\{x\in X:f(y)\in S\}.$$

If $X$ and $Y$ have been given topologies--say $\mathcal T_X$ and $\mathcal T_Y$, respectively--then we say that $f:X\to Y$ is continuous if and only if for each $U\in\mathcal T_Y$ we have $f^{-1}[U]\in\mathcal T_X.$ A few nice facts about preimages that will help for this problem (I leave them to you to verify, by definition of preimage and function):

Given any sets $X,Y$ and any function $f:X\to Y,$ we have $f^{-1}[\emptyset]=\emptyset$ and $f^{-1}[Y]=X.$

As a consequence of these facts, a function $f:X\to Y$ (where $X$ and $Y$ have been given topologies) will be continuous if and only if each non-trivial open subset of $Y$ has an open preimage under $f.$ For this particular example, this means that $f:X\to Y$ is continuous if and only if $$f^{-1}\bigl[\{A\}\bigr]\in\bigl\{\emptyset,X,\{1\},\{3,4\},\{1,3,4\}\bigr\}.$$

Now, as a rule, if we have some sets $X$ and $Y$ and some $f:X\to Y,$ and if $S$ is a proper subset of $Y,$ then knowing what $f^{-1}[S]$ is won't be enough to tell us how $f$ is defined. In fact, it will be enough if and only if $Y$ has at most two elements, and exactly one element more than $S.$ (Why? It's a nice exercise to prove this result.) Fortunately, in your particular example, $Y$ has exactly two elements, and exactly one element more than $\{A\}\subsetneq Y.$ So, determining the preimage of $\{A\}$ under $f$ is identical to determining $f.$

For instance, in order to have $f:X\to Y$ such that $f^{-1}\bigl[\{A\}\bigr]=\{1,3,4\},$ we must have $$f(x):=\begin{cases}A & \text{if }x=1,3,\text{ or }4\\B & \text{if }x=2\text{ or }5.\end{cases}$$ It should be clear that this function gives us the desired preimage, but why can't any other function $X\to Y$ give us the same preimage? Well, note that if $g:X\to Y,$ then $$g^{-1}\bigl[\{A\}\bigr]:=\bigl\{x\in X:g(x)\in\{A\}\bigr\}=\{x\in X:g(x)=A\}.$$ Thus, in order to have $g^{-1}\bigl[\{A\}\bigr]=\{1,3,4\},$ we need to know that $g(x)=A$ if and only if $x\in\{1,3,4\}.$ Knowing that $g(x)=A$ for all $x\in\{1,3,4\}$ determines $g(x)$ for most of our $x\in X,$ and knowing that $B$ is the only element of $Y$ that isn't equal to $A$ tells us that $g(x)=B$ for all $x\in X\setminus\{1,3,4\}=\{2,5\}.$ Hence, $g(x)=f(x)$ (where $f$ is as defined above) for all $x\in X.$

The other $4$ continuous functions $f:X\to Y$ are as follows:

$$f(x):=\begin{cases}A & \text{if }x=3\text{ or }4\\B & \text{if }x=1,2\text{ or }5\end{cases}$$

$$f(x):=\begin{cases}A & \text{if }x=1\\B & \text{if }x=2,3,4,\text{ or }5\end{cases}$$

$$f(x):=A\: (\text{for all }x\in X)$$

$$f(x):=B\: (\text{for all }x\in X)$$

I leave it to you to verify that $f^{-1}\bigl[\{A\}\bigr]$ is indeed open in each case.

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Actually there are five, corresponding to the five open subsets of $X$. –  Stefan Hamcke Feb 10 '14 at 16:57
@Stefan: D'oh! Thank you. Fixed. –  Cameron Buie Feb 10 '14 at 17:01
Could you specify why the 8 functions that I have listed are not the complete list of continuous functions from X to Y? I adopted the strategy of mapping each open set 9 –  kathystehl Feb 10 '14 at 17:26
@David: $A$ is not a priori a subset of $Y,$ so need not be open. In fact, $A$ need not even be a set! However, $\{A\}$ is an open subset of $Y.$ Likewise, $B$ need not be a subset of $Y$, nor even a set. However, $\{B\}$ is a closed (and non-open) subset of $Y.$ –  Cameron Buie Feb 11 '14 at 20:47
As I discussed above, given an arbitrary function $f:X\to Y,$ we have that $f$ is continuous if and only if $f^{-1}\bigl[\{A\}\bigr]$ is open. Verifying openness of the preimages of $\{A\}$ under the functions I defined above is the same as verifying those functions' continuity. –  Cameron Buie Feb 11 '14 at 20:49