8.7 Continuity and limits[2B8]
[0KB] 1 Let \((X, τ )\) and \((Y, σ)\) be two topological spaces, with \((Y, σ)\) Hausdorff. 2 Let \(E ⊆ X\) and \(f : E → Y\) . Let also \(x_ 0\) be an accumulation point of \(E\) in \(X\). We define that \(\lim _{x→x_ 0} f (x) = ℓ ∈ Y\) if and only if, for every neighborhood \(V\) of \(ℓ\) in \(Y\), there exists \(U\) neighbourhood of \(x_ 0\) in \(X\) such that \(f (U ∩ E ⧵ \{ x_ 0 \} ) ⊆ V\) .
[2B9]Let \((X, τ )\) and \((Y, σ)\) be two topological spaces, with \((Y, σ)\) Hausdorff; let \(f:X\to Y\) be a function.
It is said that \(f\) is continuous in \(x_ 0\) if \(\lim _{x→x_ 0} f (x) = f(x_ 0)\).
It is said that \(f\) is continuous if (equivalently)
if \(f\) is continuous at every point, that is \(\lim _{x→y} f (x) = f(y)\) for every \(y∈ X\), or
if \(f^{-1}(A)∈ τ\) for each \(A∈ σ\).
(Thm. 5.7.4 in the notes [ 2 ] .).
A continuous bijective function \(f:X\to Y\) such that the inverse function \(f^{-1}:Y\to X\) is again continuous, is called homeomorphism.
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[2BB](Proposed on 2022-12) Consider this statement.
«Let \(f:X\to Y\) and \(x_ 0\in X\), then \(f\) is continuous at \(x_ 0\) when, for every open set \(B\subseteq Y\) with \(f(x_ 0)\in B\), we have that \(f^{-1}(B)\) is open.»
This statement is incorrect.
Build an example of a function \(f:{\mathbb {R}}\to {\mathbb {R}}\) that is continuous at \(x_ 0=0\) but such that, for every \(J=(a,b)\) open non-empty bounded interval, \(f^{-1}(J)\) is not open. Hidden solution: [UNACCESSIBLE UUID ’2BC’]
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[225]Difficulty:*.
Let \(Y\) be a topological space. We say that \(Y\) satisfies the property (P) with respect to a topological space \(X\) when it satisfies this condition: for every dense subset \(A⊆ X\) and every pair of continuous functions \(f,g: X→ Y\) such that \(f(a)=g(a)\) for every \(a∈ A\), necessarily there follows that \(f=g\).
Prove that \(Y\) is Hausdorff if and only if it satisfies the property (P) with respect to any topological space \(X\).
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[0KC] Prerequisites:251.Explain how Definition 261 can be seen as a special case of Definition 263. (Hint. proceed as in the note 251 and set \(E=J,X=I,x_ 0=∞\)).
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[0KD]Prerequisites:3.Let \(X,Y\) be topological Hausdorff space. Let \(E⊆ X\), let \(f:E→ Y\), and suppose that \(x_ 0\) is an accumulation point of \(E\) in \(X\).
If \(\lim _{x→ x_ 0}f(x)=ℓ\) then, for each net \(φ:J→ X\) with \(\lim _{j∈ J} φ(j) = x_ 0\) we have \(\lim _{j∈ J} f(φ(j)) = ℓ\).
Consider the filtering set \(J\) given by the neighborhoods of \(x_ 0\); 3 consider nets \(φ:J→ X\) with the property that \(φ(U)∈ U⧵\{ x_ 0\} \) for each \(U∈ J\). We note that \(\lim _{j∈ J} φ(j) = x_ 0\).
If for each such net \(\lim _{j∈ J} f(φ(j)) = ℓ\), then \(\lim _{x→ x_ 0}f(x)=ℓ\).
Hidden solution: [UNACCESSIBLE UUID ’0KF’]
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[0KG]Prerequisites:1,3.Let \(X,Y\) be Hausdorff topological spaces. Let \(f:X→ Y\), \(x_ 0\in X\). The following are equivalent.
\(f\) is continuous at \(x_ 0\);
for each net \(φ:J→ X\) such that
\[ \lim _{j∈ J} φ(j) = x_ 0 \]we have
\[ \lim _{j∈ J} f(φ(j)) = f(x_ 0)\quad . \]
Hint, for proving that 2 implies 1. Suppose that \(x_ 0\) is an accumulation point. Consider the filtering set \(J\) given by the neighborhoods of \(x_ 0\); consider nets \(φ:J→ X\) with the property that \(φ(U)∈ U\) for each \(U∈ J\); note that \(\lim _{j∈ J} φ(j) = x_ 0\).
Hidden solution: [UNACCESSIBLE UUID ’0KH’]