If a socket leaves its default room, socket.to('room').emit messages will be received by it despite being sender

Describe the bug
If a socket leaves the default room which matches its socket id, socket.to('room').emit messages will be receives by the client of the sender, even though the documentation makes no mention of this.

To Reproduce

Please fill the following code example:

Socket.IO server version: 4.5.3

Server

import { Server } from "socket.io";

const io = new Server(3000, {});

io.on("connection", (socket) => {
  socket.leave(socket.id);
  OR
  socket.leaveAll();

  socket.join('TEST');

  socket.on("test", (message) => {
    socket.to('TEST').emit('test', message);
  });
});

Socket.IO client version: 4.5.3

Client

import { io } from "socket.io-client";

const socket = io("ws://localhost:3000/", {});

socket.on("test", (message) => {
  console.log(message);
  // Is logged by sender as well as other clients in the room
});

socket.emit('test', 'hello');

Expected behavior
I would not expect that leaving the default room would mean the sender receives a message that otherwise it shouldn't. My idea of leaving the default room is that I can ensure each connected client is only in a single room, and then when emitting to the room they're in I was hoping to use:

const [thisRoom] = socket.rooms;
socket.to(thisRoom).emit('test', message);

Perhaps there is a different expected way to use rooms but it would seem a bit clunky to use const [, second] = socket.rooms if I leave each client in the default room?

Platform:

• Device: Windows
• OS: Server 2019 Server, 11 Client

Start by running the provided Socket.IO 4.5.3 server and client reproduction. Trace the room membership after socket.leave(socket.id) or socket.leaveAll(), then follow socket.to('TEST').emit and inspect socket.rooms. Done means the observed sender behavior is explained and the implementation or documented behavior is corrected consistently.