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Proof

(related to Proposition: Image of a Compact Set Under a Continuous Function)

• Let \(X,Y\) be metric spaces.
• Let $f:X\mapsto Y$ a continuous function.
• Let \(C\subset X\) be a compact subset of $X$.
• We have to show that the image \(f( C)\subset Y\) is also compact.
  • Let $(U_i)_{i\in I}$ be an open cover of $f( C)$.
  • Because $f$ is continuous, it follows from the definition of continuity using open sets that the inverse images $V_i:=f^{-1}(U)$ are open in $X$.
  • Thus, $(V_i)_{i\in I}$ be an open cover of $C$.
  • Since $C$ is compact, there is a finite subcover $V_{i_1}\cup V_{i_2}\cup \ldots \cup V_{i_n}\supset C$.
  • It follows that $U_{i_1}\cup U_{i_2}\cup \ldots \cup U_{i_n}\supset f( C)$.
  • This means that $f( C)$ has a finite subcover in $Y$.
  • This means that $f( C)$ is compact in $Y$.
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References

Bibliography

1. Forster Otto: "Analysis 2, Differentialrechnung im \(\mathbb R^n\), Gewöhnliche Differentialgleichungen", Vieweg Studium, 1984