Jerry's Digital Garden

Bibliography
/Part 2: General Topology/Chapter 4: Topological Spaces/Section 4.5: Interior, Closure, Boundary

Proposition 4.5.3. Let $X, Y$ be topological spaces, $A \subset X$, and $f: X \to Y$ be continuous, then $f(\ol{A}) \subset \ol{f(A)}$.

Proof. Since $f$ is continuous, $f^{-1}(\ol{f(A)})$ is closed and contains $A$.$\square$

Direct Backlinks

  • Section 5.5: Completeness
  • Theorem 5.5.6: Extension of Uniformly Continuous Functions, [Theorem 1.3.2, Bou13]
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Jerry's Digital Garden

Bibliography

Direct Backlinks

  • Section 5.5: Completeness
  • Theorem 5.5.6: Extension of Uniformly Continuous Functions, [Theorem 1.3.2, Bou13]
Powered by Spec