# Topology Counterexample - Book Error?: Prove: Let f:(X, T) -> (Y, T*) be open and onto, and let B be a Base for T. Show f(B) is a base for T*

From Schaum's Topology Outline, Chapter 7, Supplementary problem #39.

Prove: Let $$f:(X,\mathscr{T})\rightarrow (Y, \mathscr{T*})$$ be open and onto, and let $$\mathscr{B}$$ be a base for $$\mathscr{T}$$. Then {$$f [B]:B\in \mathscr{B}$$} is a base for $$\mathscr{T*}$$.

I think this cannot be proved in general. [Edit: The book does not specify whether or not $$f$$ is continuous. Other problems in the same section do occasionally say "Let $$f$$ be continuous", so I don't think we can assume continuity here.]

Here is what I believe is a counterexample. Please let me know if I'm missing something:

Let $$X = \{a,b,c,d\}$$ and $$\mathscr{T} = \{X, \{a,b\}, \{c,d\}, \emptyset\}$$

Let $$Y = \{1,2,3\}$$ and $$\mathscr{T*} = \{Y, \{1,2\}, \{2,3\}, \{2\}, \emptyset\}$$

Define the function, $$f:X\rightarrow Y, f(a) = 1; f(b) = f(c) = 2; f(d) = 3$$

Then, $$f$$ is open because it maps open sets to open sets, and $$f$$ is onto because every element of $$Y$$ is mapped to.

Now, let $$\mathscr{B} = \{\{a,b\},\{c,d\}\}$$ be a base for $$\mathscr{T}$$.

This is base because its elements are open sets and every open set in $$\mathscr{T}$$ is a union of members of $$\mathscr{B}$$. (NB: the empty union provides the empty set)

But then $$\{f [B]:B\in \mathscr{B}\}=\{\{1,2\},\{2,3\}\}$$, which is NOT a base for $$\mathscr{T*}$$ because you cannot obtain the open set $$\{2\}$$ with unions of elements of this set.

So the general statement of the proof is false.

• Your $f$ is not continuous. I guess $f$ has to be continuous? Commented Mar 14, 2022 at 20:43
• Good point, but there's nothing in the statement of the problem that says $f$ is continuous. Other problems don't seem to assume that it is without saying so, i.e., other problems explicitly tell you "Let f be continous". So since this one doesn't, I'm not assuming it. Commented Mar 14, 2022 at 20:48
• If we assume $f$ is additionally continuous and $y \in U \subset Y$ is an open neighbourhood of $y$, then $V := f^{-1}(U)$ is open and there is a point $x \in V$ with $f(x) = y$. Because $\mathscr{B}$ is a basis there is $B \in \mathscr{B}$ with $x \in B \subset V$ open. Then $y = f(x) \in f(B) \subset f(V) = U$ is an open neighbourhood. So it seems to hold in this case. Commented Mar 14, 2022 at 20:55
• Indeed if you only want open and onto you can simplify/generify your example: Let $X$ be a set and $\mathscr{T} = \{\emptyset, X\}$ be the indiscrete topology and $\mathscr{T}' = \mathscr{P}(X)$ be the discrete topology. Then $f \colon (X, \mathscr{T}) \to (X, \mathscr{T}'),\, x \mapsto x$ is bijective and open, but a counterexample if $|X| \geq 2$, whatever basis you choose on $(X, \mathscr{T})$. So the statement doesn't make a lot of sense without $f$ being continuous. Commented Mar 14, 2022 at 21:01