Let X and Y be non-empty sets and let $\tau$ and $\nu$ be, respectively, the discrete topologies on X and Y. Every function $f: \left ( X,\tau \right )\rightarrow \left ( Y,\nu \right )$ is continuous.

Here's my attempt at the proof:

Recall that for any two topological space $\left ( X,\tau \right )$ and $\left ( Y,\nu \right )$, a function $f:\left ( X,\tau \right )\rightarrow \left ( Y,\nu \right )$ is called continuous IFF for every $u \in \nu$, we have $f^{-1}\left ( u \right )\in \tau$

Now, $\tau$ is discrete on X so $\tau=\left \{ \varnothing,\left \{ u_{i} \right \}_{i \in I} \right \}$ where $X \in \left \{ u_{i} \right \}_{i \in I}$

Also, $\nu$ is discrete topology on Y so $\nu=\left \{ \varnothing,\left \{ u_{i} \right \}_{i \in I} \right \}$ where $Y \in \left \{ u_{i} \right \}_{i \in I}$

Indeed, $f^{-1}\left ( \varnothing \right )=\varnothing \in Y$. Then, $f^{-1}\left ( u \right )=\left \{ x \in X \mid f\left ( x \right ) \in Y \right \}=X$ since the map f is between the topological space $\left ( X,\tau \right ) to \left ( Y,\nu \right ).$

Obviously, $\left \{ u_{i} \right \}_{i \in I}\setminus Y$ are subsets of Y smaller than Y itself. Since every element $u \in Y$ is mapped from X, then clearly, every element in $\left \{ u_{i} \right \}_{i \in I}\setminus Y$ are from X too. Hence, $f^{-1}\left ( \left \{ u_{i} \right \}_{i \in I}\setminus Y \right ) \in X \in \tau$.

Hence, f is continuous.

Is my proof flawed?

Any help is appreciated. Thanks in advance.

  • $\begingroup$ What do you mean with $\{u_i\}_{i\in I}$? I should say $\tau=\wp(X)$. $\endgroup$
    – drhab
    Aug 18, 2016 at 8:56
  • $\begingroup$ I meant that to be the collection of subsets of Y in $\nu$ without the empty set being included-included in $\nu$ but separately.@drhab $\endgroup$ Aug 18, 2016 at 8:57
  • $\begingroup$ What's the use of setting them apart from $\varnothing$ which is also a subset of $X$ in $\tau$? Do you agree with $\tau=\wp(X)$ (i.e. $\tau$ is the collection of all subsets of $X$)? No bothering with indices and immediately clear. $\endgroup$
    – drhab
    Aug 18, 2016 at 9:01
  • $\begingroup$ Yes I agree. I understand that in general, there is no need for separation of the empty set but it was done so to reduce confusion. I don't in particular enjoy indexing notations. $\endgroup$ Aug 18, 2016 at 9:03

2 Answers 2


The discrete topology on $X$ is not $\tau=\{\varnothing,\{u_i\}_{i∈I}\}$. While each $\{u_i\}_{i∈I}$ is indeed open, this in fact implies every subset of $X$ is open (because a union of open sets is open), so $\tau = \mathcal{P}(X)$ (the power set of $X$).

Since every subset of $X$ is open, the preimage of every set in $Y$ is open. You don't even need to assume $Y$ has the discrete topology.


I think you make things more complicated than necessary.

If $X$ is equipped with discrete topology then every function $f:X\to Z$ (where $Z$ denotes an arbitrary topological space) is continuous.

This because the preimage of every subset of $Z$ w.r.t. $f$ is an open set.

Note that these preimages are subsets of $X$ and every subset of $X$ is open. This because $X$ is equipped with the discrete topology: $\tau_X=\wp(X)$.

The looks of the topology on $Z$ is completely irrelevant.


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