How to Prove a Function is Not an Open Function

TL;DR

A simple proof shows that function f, where f(x) = x^2, is not an open mapping.

Transcript

hi everyone in this video we're going to prove that if we have the function f from the set of real numbers into the set of real numbers given by f of X equals x squared that it is not an open mapping so let me first briefly recall what it means for a mapping to be open so recall if you have a function f from a topological space capital X to a topol... Read More

Key Insights

• 🤗 Open mappings map open sets to open sets.
• 🤗 Function f, where f(x) = x^2, is being analyzed for open mapping properties.
• 🤗 To prove that f is not an open mapping, a single example is sufficient.
• 😫 Taking the set O = [-1, 1] and evaluating f on this set results in the set [0, 1].
• ✴️ The set [0, 1] is not open because it lacks an interior point.

Questions & Answers

Q: What is the definition of an open mapping?

An open mapping is a function that maps open sets to open sets, meaning the direct image of an open set is also an open set.

Q: How can we prove that function f is not an open mapping?

To prove this, we need to find one example where the direct image of an open set under f is not an open set. In this case, taking the set O = [-1, 1] and evaluating f on this set gives us the set [0, 1], which is not open.

Q: Why is [0, 1] not an open set?

For a set to be open, every point in the set needs to have an interior point within the set. In the case of [0, 1], the point 0 does not have an interior point within the set, as there are points less than 0 outside of the set.

Q: How does proving that function f is not an open mapping complete the proof?

By demonstrating that there exists at least one example where f fails to map an open set to an open set, we can conclude that f is not an open mapping.

Summary & Key Takeaways

• An open mapping is a function that maps open sets to open sets.

• To prove that function f is not open, we need to find one example where it fails.

• Taking the set O = [-1, 1], the direct image of O under f results in the set [0, 1].

• Since [0, 1] is not an open set (0 is not an interior point), it proves that f is not an open mapping.