A topological space $Y$ is a Hausdorff space, iff for every(?) two continuous functions $f,g: X\to Y$ holds, that $\{x\in X| f(x)=g(x)\}$ is closed in $X$.
I am asked to proof this statement. I hope I translated it correctly. I am not sure if it is meant that the condition really has to hold for every two continuous functions. Thats why I marked it with (?).
I tried to proof this, but did not succeed to this point and I wonder if this is even true. There is something that seems odd to me.
I know that a for a topological space it is true, that $X$ is Hausdorff, iff $\Delta=\{(x,x)|x\in X\}\subseteq X\times X$ is closed (with regards to the product topology).
And this seems related to the statement above, but much more clear.
In the question above I am not sure what $X$ is and which topology is used. Does it matter here?
Can you confirm, that this question is correct?
Thanks in advance.
It is a correct statement and in fact the "closed diagonal" characterisation ($\Delta$ is usually called the diagonal of $X$) is the means to prove it, and a special case of it as well:
Defining the equality set $E(f,g) := \{x: \in X: f(x) = g(x)\}$ for two continuous function $f: X \to Y$, we see that it equals $(f \nabla g)^{-1}[\Delta]$, where $f\nabla g: X \to Y \times Y$ defined by $(f\nabla g)(x) = (f(x), g(x))$ is continuous iff $f$ and $g$ are.
So $Y$ Hausdorff implies diagonal closed implies all equality sets closed.
The reverse is easy: if all equality sets are closed (for all domains and all pairs of continuous $f,g$), apply it to the 2 projections from $Y \times Y \to Y$ to get that $\Delta = E(\pi_1, \pi_2)$ is closed and we get the reverse almost for free:
All equality sets closed implies $E(\pi_1,\pi_2)= \Delta$ closed implies $Y$ Hausdorff.