# Show that a constant mapping between metric spaces is continuous

Let $X$ and $Y$ be metric spaces and $f$ a mapping of $X$ into $Y$.If $f$ is a constant mapping, show that $f$ is continuous. Use this to show that a continuous mapping need not have the property that the image of every open set is open.

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What have you tried? –  Jason DeVito Apr 18 '12 at 1:46
Is this hw? What have you tried. etc, etc, etc.... –  john w. Apr 18 '12 at 1:46
What is the image of such an $f$ (that is, the image of the open subset $X$ of $X$)? Is that always open in $Y$? –  Dylan Moreland Apr 18 '12 at 1:48
I think you have to use this result. If $f$ is continuous then inverse image of open set is open. –  Kns Apr 18 '12 at 2:42
It's not even necessary to look preimages of open sets. From the non-negativity of the metric it follows that constant maps are $1$-Lipschitz. –  Thomas E. Apr 19 '12 at 7:37
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Let $f: X \to Y$ be a constant map, that is, $f(x) = c$ for all $x$ in $X$.

By definition, a function is continuous if for $O \subset Y$ open you have that $f^{-1} (O)$ is open in $X$.

So let $O$ be an open set in $Y$. Then there are two cases that can happen:

(i) $c \in O$. Then $f^{-1} (O) = X$. $X$ is the whole space and this is an open set.

(ii) $c \notin O$. Then $f^{-1}(O) = \varnothing$ which is also an open set.

Hence we have shown that if $f$ is constant then its preimages of open sets are open or in other words: $f$ is continuous.

To see that a continuous function need not map open sets to open sets consider $f: \mathbb R \to \mathbb R$, $x \mapsto x^2$. Then $(-1,1)$ is open in $\mathbb R$ with the standard topology but $f((-1,1)) = [0,1)$ is not open.

Hope this helps.

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Suppose there is a $c\in Y$ and we define $f(x)=c$ for all $x\in X$.

What is the preimage of an open set containing $c$? Not containing $c$?

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A function $f$ is continuous if and only if the inverse image of any open set of $Y$ is open in $X$ (Kunjan shah suggested the "only if" part, but you need the "if" part), thus you just have to think about this inverse images. In fact you don't even need $X$ and $Y$ to be metric, just topological spaces. To get a continuous map which does not preserve open sets, think about $X=\mathbb{R}=Y$ endowed with the usual topology.

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Constant maps are $1$-Lipschitz and hence continuous. Let $f:(X,d)\to (Y,e)$ be a constant map to some $y\in Y$, i.e. $f\equiv y$.
Then $e(f(x_{1}),f(x_{2}))=e(y,y)=0\leq d(x_{1},x_{2})$ for all $x_{1},x_{2}\in X$ by non-negativity of the metric $d$.
You might want not to use $y$ twice. –  nik Apr 19 '12 at 7:56