A function is continuous if and only if the pre-image of every basis element is open

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Important for functional analysis.

Work required:

  1. Extend to the "[ilmath]\epsilon[/ilmath]-[ilmath]\delta[/ilmath]" form of continuity, with metric spaces. That is after all an instance of this
  2. Expand on basis generated by
  3. Expand on topology induced by a metric
  4. Make sure they're al "united"


Let [ilmath](X,\mathcal{ J })[/ilmath] and [ilmath](Y,\mathcal{ K })[/ilmath] be topological spaces, let [ilmath]\mathcal{B} [/ilmath] be a basis of [ilmath](Y,\mathcal{ K })[/ilmath], and let [ilmath]f:X\rightarrow Y[/ilmath] be a mapping between them[Note 1]. Then[1]:

  • [ilmath]f:X\rightarrow Y[/ilmath] is continuous if and only if [ilmath]\forall B\in\mathcal{B}[f^{-1}(B)\in\mathcal{J}][/ilmath]


Note that for [ilmath]f:X\rightarrow Y[/ilmath] to be continuous means:

  • [ilmath]\forall U\in\mathcal{K}[f^{-1}(U)\in\mathcal{J}][/ilmath]

We claim that we have this if and only if:

  • [ilmath]\forall B\in\mathcal{B}[f^{-1}(B)\in\mathcal{J}][/ilmath]

If this is true we have found a tool that lets us solve the problem of continuity by considering a much "smaller" collection of sets, often with some nice properties. For example the open balls of a metric space rather than any open set

This should also be fairly easy to prove, as [ilmath]\mathcal{B} [/ilmath] is a topological basis of [ilmath](Y,\mathcal{ K })[/ilmath] if (and only if - see definitions and iff):

  1. [ilmath]\forall B\in\mathcal{B}[B\in\mathcal{K}][/ilmath] - the basis elements themselves are open, and
  2. [ilmath]\forall U\in\mathcal{K}\exists(B_\alpha)_{\alpha\in I}[\bigcup_{\alpha\in I}B_\alpha\eq U][/ilmath] - every open set is the union of some arbitrary collection of basis elements

Number one means that to prove continuity by definition we'd have to deal with the basis elements anyway and number two fits in nicely as:

As we will see in the proof.


There are 2 parts to this proof:

  • [ilmath]\forall B\in\mathcal{B}[f^{-1}(B)\in\mathcal{J}]\implies\forall U\in\mathcal{K}[f^{-1}(U)\in\mathcal{J}][/ilmath] and
  • [ilmath]\forall B\in\mathcal{B}[f^{-1}(B)\in\mathcal{J}]\impliedby\forall U\in\mathcal{K}[f^{-1}(U)\in\mathcal{J}][/ilmath]

our claim is on the LHS and continuity is on the right.

Proof: [ilmath]\implies[/ilmath]

Suppose [ilmath]\forall B\in\mathcal{B}[f^{-1}(B)\in\mathcal{J}][/ilmath], show that [ilmath]f[/ilmath] is continuous in this case.

  • Let [ilmath]U\in\mathcal{K} [/ilmath] be given - we wish to show that [ilmath]f^{-1}(U)\in\mathcal{J} [/ilmath]
    • As [ilmath]\mathcal{B} [/ilmath] is a basis of [ilmath](Y,\mathcal{ K })[/ilmath], thus we know that:
      • [ilmath]\exists(B_\alpha)_{\alpha\in I}\subseteq\mathcal{B}[\bigcup_{\alpha\in I}B_\alpha\eq U][/ilmath]
    • Choose [ilmath]\{B_\alpha\}_{\alpha\in I}\subseteq\mathcal{B} [/ilmath] to be a family of basis elements such that [ilmath]U\eq\bigcup_{\alpha\in I}B_\alpha[/ilmath]
      • Note that [ilmath]f^{-1}(U)\eq f^{-1}(\bigcup_{\alpha\in I}B_\alpha)[/ilmath] as [ilmath]U\eq\bigcup_{\alpha\in I}B_\alpha[/ilmath]
      • But by properties of the pre-image of a function we see:
        • [ilmath]f^{-1}(U)\eq f^{-1}(\bigcup_{\alpha\in I}B_\alpha)\eq\bigcup_{\alpha\in I}f^{-1}(B_\alpha)[/ilmath]
        • By hypothesis [ilmath]\forall B\in\mathcal{B}[f^{-1}(B)\in\mathcal{J}][/ilmath], so as the [ilmath]B_\alpha\in\mathcal{B} [/ilmath] we see:
          • [ilmath]\forall\alpha\in I[f^{-1}(B_\alpha)\in\mathcal{J}][/ilmath]
          • Since [ilmath]\mathcal{J} [/ilmath] is a topology it is closed under arbitrary union.
            • We have shown each element of the union, a set of the form: [ilmath]f^{-1}(B_\alpha)\in\mathcal{J} [/ilmath]
              • So [ilmath]\bigcup_{\alpha\in I}f^{-1}(B_\alpha)\in\mathcal{J} [/ilmath]
                • But: [ilmath]f^{-1}(U)\eq f^{-1}(\bigcup_{\alpha\in I}B_\alpha)\eq\bigcup_{\alpha\in I}f^{-1}(B_\alpha)[/ilmath]
                  • So [ilmath]f^{-1}(U)\in\mathcal{J} [/ilmath]
  • Since [ilmath]U\in\mathcal{K} [/ilmath] was arbitrary we have shown [ilmath]f^{-1}(U)\in\mathcal{J} [/ilmath] for all, the very definition of continuity

Proof: [ilmath]\impliedby[/ilmath]

Suppose that [ilmath]f:X\rightarrow Y[/ilmath] is continuous. We want to show [ilmath]\forall B\in\mathcal{B}[f^{-1}(B)\in\mathcal{J}][/ilmath]

  • Let [ilmath]B\in\mathcal{B} [/ilmath] be given
    • Then [ilmath]B\in\mathcal{K} [/ilmath] as [ilmath]\mathcal{B} [/ilmath] is a basis of [ilmath](Y,\mathcal{ K })[/ilmath]
      • By continuity of [ilmath]f[/ilmath]: [ilmath]\forall U\in\mathcal{K}[f^{-1}(U)\in\mathcal{J}][/ilmath], thus:
        • [ilmath]f^{-1}(B)\in\mathcal{J} [/ilmath] - as required.
  • Since [ilmath]B\in\mathcal{B} [/ilmath] was arbitrary, we have shown it for all, as required.


  1. We do not assume anything about this mapping, for example it need not be continuous


  1. Introduction to Topological Manifolds - John M. Lee