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I read the definition of a continuous function between topological spaces a lot of times, but I'm having difficulties to apply it to a simple example.

Given two topological spaces $(X,\tau_1)$ and $(Y,\tau_2)$, i.e.

$$ X=\{a,b,c\}, \tau_1=\{X,\emptyset,\{a,b\},\{c\},\{b,c\},\{b\} \} $$ $$ Y=\{d,e,f\}, \tau_2=\{Y,\emptyset,\{d,e\},\{f\},\{e,f\},\{e\} \} $$

What would be an example of a continuous function (if any exists) $f:X \rightarrow Y$?

Wiki says:

A function f : X→ Y between topological spaces is called continuous if for all x ∈ X and all neighbourhoods N of f(x) there is a neighbourhood M of x such that f(M) ⊆ N.

Here I have trouble understanding what is meant by this definition. Does $x \in X $ refer to elements of my topology $\tau_1$/ $\tau_2$ or elements of $X$/$Y$?

Let's say I map elements of $\tau_1$ to elements of $\tau_2$ like this: $$ \{a,b,c\}\rightarrow \{d,e,f\}$$ $$ \{a,b\}\rightarrow \{d,e\} $$ $$\{c\} \rightarrow \{f\}$$ $$\{b,c\}\rightarrow\{e,f\}$$ $$\{b\}\rightarrow\{e\} $$

Would this be correct? And how can I check if this mapping is continuous? Or in other words, what is a neighborhood of a set in the topologies, e.g. what is a neighborhood of $\{a,b,c\}$ or a neighborhood of $ \{e\} $?

Well, I'm quite confused about all this, so a simple example would be nice to help me get less confused :).

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  • $\begingroup$ Well, this is a dumb example perhaps, but the identity map from $X$ to $X$ (endowed with the same topology for domain and image) is continuous. $\endgroup$ – Chill2Macht May 21 '16 at 18:20
  • $\begingroup$ I don't see how you can be confused about the notation $x\in X$ : isn't it perfectly clear that $x$ is supposed to be an element of $X$ ? $\endgroup$ – Captain Lama May 21 '16 at 18:20
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What you gave is the definition for a function to be continuous at a point. The usual definition for a function to be continuous (which is equivalent with $f$ being continuous at every point) is that

$f$ is continuous if the inverse image of every open set is open.

It is easy to see, by the way you constructed your example, that indeed it is continuous.

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  • $\begingroup$ Well, this definition I certainly understand :). But is it possible to define a function based on the definition above as well with regard to my example? $\endgroup$ – holistic May 21 '16 at 19:59
  • $\begingroup$ Sure, if you take as the definition that $f$ is continuous that it is continuous at each point. For each point in $Y$, look at the inverse images of each open set containing that point and check that they are open in $X$. $\endgroup$ – user333870 May 21 '16 at 20:06
  • $\begingroup$ Sorry, that's what I don't get somehow :(. Can you maybe give a simple example based on the above sets (if possible)? $\endgroup$ – holistic May 21 '16 at 20:27
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    $\begingroup$ Take, for example, the element $e \in Y$. All of the open sets containing $e$ are the sets $Y$, $\{e\}$, $\{e,f\}$ and $\{d,e\}$. The inverse images of each of these sets are, respectively, $X$, $\{b\}$, $\{b,c\}$, and $\{b,c\}$, which are all open sets in $X$. By the way, you should define the function on $X$, not on the open sets. That is, the function $f: X \rightarrow Y$ is the function that maps $f(a) = d, f(b) = e,$ and $f(c) = f$. $\endgroup$ – user333870 May 21 '16 at 20:31
  • $\begingroup$ Got it :), thank you very much! $\endgroup$ – holistic May 21 '16 at 20:43

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