Does a continuous function preserve Hausdorff or metrizable properties? [closed]

Prove/Disprove- If $f: X\to Y$ is a continuous function and $X$ is Hausdorff, then $Y$ is Hausdorff. Can someone provide a counterexample for this? In other words, show by counterexample that Hausdorff-ness is not preserved under continuous functions.

Also, what would be a counterexample for X being metrizable, Y not metrizable?

closed as off-topic by Moishe Kohan, user99914, Rebellos, Shailesh, AquaDec 7 '17 at 12:15

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• Consider the embedding of the real line into the the line with two origins. – Levent Dec 6 '17 at 21:35
• Probably easiest to disprove if $f$ is not onto. Then you can choose $Y$ as $X\sqcup Z$ where $Z$ is not Hausdorff, for example. If $f$ is onto, going to need more work, but I think $\mathbb R\to\mathbb R/\mathbb Q$ might be a counterexample. – Thomas Andrews Dec 6 '17 at 21:35
• Take any non-Hausdorff space containing at least one point and consider the embedding of this point into the space. – Levent Dec 6 '17 at 21:36
• Consider the trivial topology for $Y$. – celtschk Dec 6 '17 at 21:38
• It is bad form to edit in another question after people have answered the original question. You should revert the question to its original form and ask about metrizability in another post. – Alex Provost Dec 6 '17 at 21:41

Hint: pick your favorite Hausdorff $X$, non-hausdorff $Y$, and a constant map $f:X \to Y$. Even if $f$ is onto or even a quotient map, there are well-known counterexamples, see e.g. Quotient Space of Hausdorff space.

Take any Hausdorff space $X,$ any non-Hausdorff topological space $Y,$ and a constant map $f:X\to Y.$ Then $f$ is continuous.

• Didn't realize it was that simple. Thanks! – FrancoM Dec 8 '17 at 3:42

If the topology on $X$ is metrizable, i.e. it comes from a metric $\mathrm{d}: X \times X \to \mathbb{R}$, then $X$ will be Hausdorff because every two different points $x,y \in X$ can be separated by open balls of radius less than $\frac{\mathrm{d}(x,y)}{2}$.

If Hausdorffness is not preserved under continuous functions, as it has been shown by others, then metrizability will not be preserved either.

The counter-examples will be similar.

Let $\mathrm{id}: (\mathbb{R}, \mathrm{Euclidean}) \to (\mathbb{R},\{0,\mathbb{R}\})$ be the identity function on the real numbers with the Euclidean topology (coming from the Euclidean metric) in the domain and the trivial topology in the co-domain. The domain is of course metrizable, the function is continuous because the pre-images of $\emptyset$ and $\mathbb{R}$ are open in the domain, but $\mathbb{R}$ with the trivial topology is not Hausdorff (it just doesn't have enough open sets for being Hausdorff. You can't separate $0$ from $1$ in the trivial topology!) and therefore, it is not metrizable.

• Thanks very much. Helped me to understand Hausdorff, metrizability, and the relationship between everything in this post. – FrancoM Dec 8 '17 at 3:41
• @FrancoM: You're welcome. – stressed out Dec 8 '17 at 8:27

If $Y$ is a non-metrisable space or a non-Hausdorff space, let $X$ be the same set but then in the discrete topology, and let $f(x) = x$. Then $f$ is continuous, as any function on a discrete space and $X$ is a metrisable space. This gives a counterexample for any property $P$ such that there is a space $Y$ that does not have $P$ and such that the discrete space of the same size does have $P$.

Some classic properties that are preserved by continuous maps: compactness, Lindelöfness, countable compactness (in the cover sense), separability, having a countable network, connectedness, path-connectedness. The above gives a counterexample for $T_0, T_1 ,\ldots, T_6$, metrisability, paracompactness, zero-dimensionality, total disconnectedness, etc.

• Thank you of helping me to tie all of these topics together. Very helpful. – FrancoM Dec 8 '17 at 3:41