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I'm given the following problem on a chapter about Topological notions:

Let $X$ be a topological space, show that there exists a continous map $X \to \{0,1\} \iff \text {X is disconnected}$, with $\{0,1\}$ given the discrete topology.

I can understand that if there is such a continous function, then the sense of "disconnection" in $\{0,1\}$ must be in $X$, but I have no experience working with these notions, and how to link the propierty of continuity (the image of open sets in $X$ is open) to this case. What else can intuitively be said about this problem?

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    It is the preimages of open sets that are open under a continuous map, whence give you a partition of $X$ into two open sets (when the map is onto). – 2017-01-31
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    I think that the map $X \to \{0,1\}$ needs to be continuous AND surjective – 2017-01-31
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    As I just checked in Lee, the mapping must be nonconstant. It is exercise 4.4 for reference. – 2017-01-31
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    @TheGeekGreek I forgot to write that, sorry. What happens if it's constant? – 2017-01-31
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    @HyperbolicMarraquetoid Then the equivalence is wrong. A constant mapping is always continuous no regard of the properties of the topological space. – 2017-01-31

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If $X$ is disconnected then there exist non-empty open sets $U,V\subset X$ such that $U\cap V=\emptyset$ and $X=U\cup V$. Let $f$ be equal to $0$ on $U$ and $1$ on $V$.

For the other direction suppose you have such a map $f$ (needs to be non-constant i.e. attains both values $0$ and $1$) and let $U$ the preimage of $0$ and $V$ the preimage of $1$. Since $\{0,1\}$ has the discrete topology the singletons are open hence $U,V$ are open sets in $X$. Then $U\cap V=f^{-1}(\{0\}\cap\{1\})=\emptyset$. Morever $X=f^{-1}(\{0,1\})=U\cup V$. Namely, $X$ is disconnected.

Since the definition that OP has is that $X$ is disconnected if-f there exists a closed and open set$ B\neq \emptyset, X$ we can show that this definition is equivalent to $X$ is disconnected if it is the disjoint union of two non-empty sets.

$(\Rightarrow)$ $X=B\cup B^c$ where $B^c$ is the complement of $B$. Then $B,B^c$ are non-empty disjoint open sets whose union is $X$.

$(\Leftarrow)$ Let $U,V$ disjoint open sets such that $X=U\cup V$. Then $V$ is the complement of $U$. Hence $U$ is open and closed and since $U,V$ non-empty we have also that $U$ is not $\emptyset$ or $X$.

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    Thank you, can I ask why your first statement is true? I mean, formally. – 2017-01-31
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    @HyperbolicMarraquetoid $X$ is disconnected if it is the union of two disjoint nonempty open sets. Here is the definition I used: https://en.wikipedia.org/wiki/Connected_space – 2017-01-31
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    Oh, I was given the definition that $X$ is open iff there exists a non-trivial clopen set. – 2017-01-31
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    @HyperbolicMarraquetoid The two definitions are equivalent. I will edit my answer. – 2017-01-31
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Let $ X $ be disconnected . Then there exists open sets $ U $ and $ V $ such that $ X=U \cup V $ and $ U \cap V = \phi $ , empty set. Let us define $ f(x)= \{0 , \ if \ x \in U \ and \ 1 \ , if \ x \in V \} $ . Then $ f: X \rightarrow \{0,1\} $ is continuous because here open sets are $ \{\ 0 \}, \{ 1\}, \{ 0,1\} $ and $ \{ \phi \} $. And so $ f^{-1} (\{0\})=U \in \tau $ , topology, $$ $$ Also $ f^{-1}(\{ 1\})=V \in \tau $ , and $ f^{-1}(\{0,1\})=X \in \tau $. $$ $$ Hence f is continuous surjection.