Kronecker "delta" function is generally defined as $\delta(i,j)=1$ if $i$ is equal to $ j$, otherwise $0$.

How about if $j$ is not an integer? I mean let $j$ is a half open interval defined as $j=(0,1]$ and $ i$ has any value on interval $[0,1]$, then can we use Kronecker delta to find if $i$ belongs to $j$? In other words, can we define like: $\delta(i,j)=1$ if $i$ is in $j$, $0$ otherwise?

What you are looking for is the indicator function: $$ \mathbf{1}_{A}(x)=\begin{cases} 1 & \text{if }x\in A\\ 0 & \text{otherwise} \end{cases} $$ In your case, $A=(0,1]$ (I have used $x$ instead of $i$ and $A$ instead of $j$ since it is more customary to use lower case letters at the end of the alphabet for real numbers and upper case letters for sets).

Note, as asmeurer points out, that the Kronecker delta $\delta_{i,j}$ can be written in terms of the indicator function: $\delta_{i,j}=\mathbf{1}_{\{j\}}(i)=\mathbf{1}_{\{i\}}(j)$.


That having been said, the answer to your original question is yes, since there is nothing stopping you from defining $\delta_{i,j}=\mathbf{1}_{j}(i)$ when $j$ is a subset of the real numbers and $i$ is a real number, but this might confuse your readers, and thus I strongly recommend against it.

  • Worth pointing out that $\delta_{i,j}$ is a special-case of the indicator function, $\mathbf{1}_{\{j\}}(i)$. – asmeurer Jun 19 '16 at 3:36

No, the convention is that the Kronecker delta tests for equality and not for some other relation such as set membership. Also the elements tested for equality are most often numbers, and even integers, but in a stretch you could use it to test equality of objects of other types. So using the Kronecker delta with one index a number and the other an interval goes against the original definition of the symbol. (This is compounded by the confusion caused by having the letter $j$ denote an interval rather than a number.)

However there is an easy notation generalising the Kronecker $\delta$ that allows you to test all kinds of relations, including set membership, which is the Iverson bracket: the convention that by enclosing any condition in brackets one obtains an expression that denotes $1$ if the condition holds, and $0$ if it fails. So instead of $\delta_{i,j}$ you can then write $[i=j]$ which makes the equality test explicit, but you can equally well use $[i\leq j]$ for an "upper triangular" relation, or if you want to test membership of an interval $I$ such as $(0,1]$ (or indeed any set) write $[i\in I]$. Personally I find this much more readable than using one of the many notations $\mathbf1_I(i)$, $\mathbf I_I(i)$, $\chi_I(i)$ for an indicator function that are in use, and also avoids monopolising yet another symbol for a limited special purpose.

To my regret the use of Iverson bracket is not yet very common in mathematical texts, but it certainly deserves to be used where it can (combinatorics for instance provides many potential uses). Since the notation introduces a logical relation in the middle of a formula, it takes a little getting used to. But this my be facilitated by typographically making the brackets stand out a bit with respect to other uses of brackets and parentheses, for instance making them bold and/or giving them additional spacing; however even without it does not really cause any conflict with other notations.

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Not the answer you're looking for? Browse other questions tagged or ask your own question.