Question

Prove that \(A\) is connected iff there is no continuous map $$ f: A \underset{\text { onto }}{\longrightarrow}\\{0,1\\} .^{5} $$ [Hint: If there is such a map, Theorem 1 shows that \(A\) is disconnected. (Why?) Conversely, if \(A=P \cup Q(P, Q\) as in Definition 3), put \(f=0\) on \(P\) and \(f=1\) on \(Q\). Use again Theorem 1 to show that \(f\) so defined is continuous on \(A\).]

Short answer

A is connected iff no continuous surjective map exists from A to \{0,1\}.

Step-by-step solution

Understand the Definitions

A space is connected if it cannot be divided into two disjoint non-empty open sets. A space is disconnected if it can be expressed as the union of two non-empty disjoint open sets. A function is continuous if the preimage of every open set is open.

Apply Theorem 1

Theorem 1 states that a space is disconnected if there exists a continuous surjective map from the space onto the discrete space \(\{0,1\}\). This is because such a map would essentially split the space into two disjoint open sets, one mapping to 0 and the other to 1.

Prove \(A\) is Disconnected \(\implies\) There Exists a Continuous Map

Assume that \(A\) is disconnected. By definition, there exist disjoint non-empty open sets \(P\) and \(Q\) such that \(A = P \cup Q\). Define a map \(f: A \rightarrow \{0,1\}\) where \(f(x) = 0\) if \(x \in P\) and \(f(x) = 1\) if \(x \in Q\). This map is surjective and continuous since the preimage of an open set in \(\{0,1\}\) is open. Hence, the existence of such a map implies \(A\) is not connected.

Prove No Continuous Map \(\implies\) \(A\) is Connected

Assume that there is no continuous surjective map from \(A\) onto \(\{0,1\}\). If \(A\) were disconnected, such a map could be constructed as shown in Step 3. Therefore, the absence of such a map means \(A\) cannot be disconnected and thus must be connected.

Key concepts

Continuous Maps

In mathematics, a continuous map is a function between two topological spaces that preserve the topological structure. This means that the image of any open set under the function is open in the image space. Intuitively, if you can draw the graph of the function without lifting your pen off the paper, the function is continuous. To dive deeper, imagine a rubber sheet being stretched or deformed; the continuous map denotes there's no tearing or gluing of the sheet.

For a function \(f: A \to B\) to be continuous, it requires that for any open set \(V\) in \(B\), its preimage \(f^{-1}(V)\) in \(A\) should also be open. This preserves the structure and ensures stability.

• If you take an open interval in \(B\), its preimage in \(A\) will also be open.
• This property is crucial when dealing with problems involving limits or convergence since it ensures the consistency of structure during transformation between spaces.

Disconnected Spaces

Disconnected spaces contrast with the concept of connected spaces in topology. A space is disconnected if it can be split into two or more non-empty, disjoint open sets. These sets cover the space fully, meaning every point in the space belongs to one of these open sets. An intuitive way to grasp this is to visualize a space as two separate islands; if you're standing on one island, you can't cross over to the other without leaving your current space.

For instance, consider the real line. A specific split, such as breaking it into \((-fty,0)\cup (0,+fty)\), clearly shows a disconnected space. There's no single continuous path from a point in one subset to a point in another.

• Disconnected spaces have non-trivial partitions that highlight breaks in the continuity of a space.
• If found, these partitions signify the failure of the space to maintain a unified topological identity.

Theorem 1 from Mathematical Analysis

Theorem 1 in the context of topology often deals with the characterization of connected versus disconnected spaces. Specifically, it links the condition of being disconnected with the ability to map a space continuously and onto the discrete space \(\{0,1\}\).

The theorem posits that if there exists a continuous surjective map from space \(A\) onto \(\{0,1\}\), then \(A\) is disconnected. This is a powerful tool because it provides a method to prove disconnection by finding or showing the absence of such a map.

• Having this map allows us to "color" parts of \(A\) differently, signifying distinct open parts.
• Conversely, if you can prove no such map exists, it shows the space is robust against being split into separate entities.

Thus, Theorem 1 not only provides theoretical proof but also offers a constructive way to analyze topological properties.